Answers
.. A = 2*(20 yd)^2 +(π/4)*(20 yd)^2
.. = (2 +π/4)*(400 yd^2)
.. = (800 +100π) yd^2
.. ≈ 1114.16 yd^2
The perimeter is that of a 20 yd circle and 80 yd more.
.. P = π*20 yd + 80 yd
.. ≈ 142.83 yd
Related Questions
The pressure P (in pounds per square foot), in a pipe varies over time. Ten times an hour, the pressure oscillates from a low of 40 to a high of 280 and then back to a low of 40. The pressure at time t = 0 is 40. Let the function P = f(t) denote the pressure in pipe at time t minutes. Find the formula for the function P=f(t),
Answers
To model the pressure function in the pipe that oscillates between 40 and 280 ten times an hour, we can use a sine function. The formula is P = f(t) = 120 sin(π/3 t) + 160.
Explanation:The pressure in the pipe oscillates between 40 and 280 ten times an hour, this is a trigonometric function scenario. Assuming the oscillation is sinusoidal, we can use a sine function to model the pressure in the pipe. The oscillation's period is 6 minutes because the pressure changes happen 10 times per hour. Thus, the function modelling this pressure will be of the form
P = a sin(b(t - c)) + d.
Given that the middle value of the pressure (between the max of 280 and the minimum of 40) is 160, this makes
'd' = 160.
The amplitude 'a' is half the total swing of the pressure which is 120.
To find 'b', we use the fact that the period of a sinusoid in this form is (2π/b).
As our period is 6 minutes, that makes 'b' = π/3.
The pressure is at a minima at t=0 so the phase shift 'c' = 0
Hence the formula P = f(t) = 120 sin(π/3 t) + 160.
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HELP PLEASE! FAST!!
1.Quadrilateral ABCD is inscribed in this circle.
What is the measure of angle B?
2. Quadrilateral ABCD is inscribed in this circle.
What is the measure of angle A?
3. Quadrilateral ABCD is inscribed in this circle.
What is the measure of angle C?
IF YOU DO NOT KNOW, DO NOT ANSWER JUST FOR POINTS. YOU WILL BE REPORTED.
Answers
x + 4x - 20 = 180
5x - 20 = 180
5x = 200
x = 40
2) We will use angle B and D to find the value of x first.
148 + x = 180
x = 32
now we will substitute this x in the value of angle A.
2x + 1 = A
2(32) + 1 = A
65 = A
3) First we will find the value of x. For that we will use the angles B and D.
x + 10 + x + 24 = 180
2x + 34 = 180
2x = 146
x = 73
So the value of x is 73. We can use that to find angle A.
x + 15 = A
73 + 15 = A
88 = A
Now we can find angle C because A and C are supplementary due to the inscribed angle theorem.
180 - 88 = C
92 = C
Hope it helps :)
(1) The measure of angle B is [tex]140^o.[/tex]
(2) The measure of angle A is [tex]65^o.[/tex]
(3) The measure of angle C is [tex]80.5^o.[/tex]
(1) The quadrilateral [tex]\(ABCD\)[/tex] is inscribed in a circle. For any quadrilateral inscribed in a circle, the opposite angles are supplementary (i.e., their sum is [tex]\(180^\circ\)).[/tex]
In the first image:
[tex]- \( \angle DAB = x^\circ \)\\ - \( \angle DCB = (4x - 20)^\circ \)[/tex]
Since these two angles are opposite angles of the inscribed quadrilateral, we have:
[tex]\[ x + (4x - 20) = 180 \][/tex]
Solving for [tex]\(x\):[/tex]
[tex]\[ 5x - 20 = 180 \][/tex]
[tex]\[ 5x = 200 \][/tex]
[tex]\[ x = 40 \][/tex]
Therefore, [tex]\( \angle B = (4x - 20) = 4(40) - 20 = 160 - 20 = 140^\circ \).[/tex]
(2) In the second image:
[tex]- \( \angle ADC = x^\circ \)\\ - \( \angle ABC = 148^\circ \)[/tex]
These are opposite angles of the inscribed quadrilateral. Thus:
[tex]\[ x + 148 = 180 \][/tex]
Solving for [tex]\(x\):[/tex]
[tex]\[ x = 180 - 148 = 32 \][/tex]
Therefore, [tex]\( \angle A = (2x + 1) = 2(32) + 1 = 64 + 1 = 65^\circ \).[/tex]
(3) In the third image:
[tex]- \( \angle DAB = (x + 15)^\circ \)\\ - \( \angle DCB = (x + 10)^\circ \)\\ - \( \angle BCD = (x + 24)^\circ \)[/tex]
Using the property that opposite angles are supplementary:
Opposite angles are [tex]\( (x + 15) \)[/tex] and [tex]\( (x + 24) \),[/tex] thus:
[tex]\[ (x + 15) + (x + 24) = 180 \][/tex]
Solving for [tex]\(x\):[/tex]
[tex]\[ 2x + 39 = 180 \][/tex]
[tex]\[ 2x = 141 \][/tex]
[tex]\[ x = 70.5 \][/tex]
Therefore, the measure of angle C is [tex]\( (x + 10) = 70.5 + 10 = 80.5^\circ \).[/tex]
Simplify completely. square root of 18y^10
Answers
3 y5 • sqrt(2)
A drama club is planning a bus trip to New York City to see a Broadway play. The table represents the cost per person for the bus rental compared to the number of people going on the trip. What function models the data, and how much per person will it cost if 12 students go on the trip?
Number of Students(n) - Cost per Student(c)
3 - 24$
6 - 12$
9 - 8$
16 - $4.5
A. n/c = 72, $12
B. nc = 9, $10
C. nc = 72, $6
D. n/c = 9, $12,
Answers
Cost function: [tex]\( nc = 72 \)[/tex]. Cost per person for 12 students: $6. Answer: C.
To determine the function that models the data and to find the cost per person if 12 students go on the trip, we need to analyze the relationship between the number of students (n) and the cost per student (c).
Given the data:
- When [tex]\( n = 3 \), \( c = 24 \)[/tex]
- When [tex]\( n = 6 \), \( c = 12 \)[/tex]
- When [tex]\( n = 9 \), \( c = 8 \)[/tex]
- When [tex]\( n = 16 \), \( c = 4.5 \)[/tex]
We can observe that as the number of students increases, the cost per student decreases. This suggests an inverse relationship between the number of students and the cost per student. The form of an inverse relationship can be expressed as:
[tex]\[ c = \frac{k}{n} \][/tex]
where [tex]\( k \)[/tex] is a constant.
To find the constant [tex]\( k \)[/tex], we can use one of the data points. Let's use the first data point ([tex]\( n = 3 \), \( c = 24 \)[/tex]):
[tex]\[ 24 = \frac{k}{3} \][/tex]
Solving for [tex]\( k \)[/tex]:
[tex]\[ k = 24 \times 3 = 72 \][/tex]
So the function that models the data is:
[tex]\[ c = \frac{72}{n} \][/tex]
Now, we need to find the cost per person if 12 students go on the trip. We substitute [tex]\( n = 12 \)[/tex] into the function:
[tex]\[ c = \frac{72}{12} = 6 \][/tex]
Therefore, the cost per person if 12 students go on the trip is $6.
The correct answer is:
C. [tex]\( nc = 72 \)[/tex], $6
To confirm this, we can check that this function fits all the provided data points:
1. For [tex]\( n = 3 \)[/tex]:
[tex]\[ c = \frac{72}{3} = 24 \][/tex] (matches the given cost)
2. For [tex]\( n = 6 \)[/tex]:
[tex]\[ c = \frac{72}{6} = 12 \][/tex] (matches the given cost)
3. For [tex]\( n = 9 \)[/tex]:
[tex]\[ c = \frac{72}{9} = 8 \][/tex] (matches the given cost)
4. For [tex]\( n = 16 \)[/tex]:
[tex]\[ c = \frac{72}{16} = 4.5 \][/tex] (matches the given cost)
Hence, the function [tex]\( c = \frac{72}{n} \)[/tex] is validated by all the data points.
1. What is the sum or difference?
4x^10 - 9x^10 (1 point)
(A). -5x^10
(B). -5x^20
(C). -36x^20
(D). -36x^20
2. What is the sum or difference?
6y^5 - 9y^5 (1 point)
(A). -3y^10
(B). 15y^5
(C). -54y^5
(D). -3y^5
3. Write the Polynomial in standard form. Then name the Polynomial based on its degree and number of terms.
2 - 11x^2 - 8x + 6x^2 (1 point)
(A). -5x^2 - 8x + 2; quadratic trinomial
(B). -5x^2 - 8x; quadratic binomial
(C). -6x^2 - 8x - 2; cubic polynomial
(D). 6x^2 - 8x + 2; cubic trinomial
4. A biologist studied the populations of white-sided jackrabbits and black-tailed jackrabbits over a 5-year period. The biologist modeled the populations, in thousands, with the following polynomials where x is time, in years.
White-sided jackrabbits: 5.5x^2 - 9.2x + 6.9
Black-tailed jackrabbits: 5.5x^2 + 9.9x + 1.3 (1 point)
(A). 11x^2 + 0.7x + 8.2
(B). 11x^2 - 0.7x + 8.2
(C). 11x^2 - 0.7x - 8.2
(D). -11x^2 + 0.7x - 8.2
Someone please help! Unit 3 Lesson 9, Polynomials and Factoring!
Answers
Solving question 1 : What is the sum or difference?
[tex] 4x^{10} - 9x^{10} \\\\
(4-9)x^{10} \\\\
-5x^{10} [/tex]
Hence, option A is correct i.e. [tex] -5x^{10} [/tex].
Solving question 2 : What is the sum or difference?
[tex] 6x^{5} - 9x^{5} \\\\
(6-9)x^{5} \\\\
-3x^{5} [/tex]
Hence, option D is correct i.e. [tex] -3x^{5} [/tex].
Solving question 3 : Write the Polynomial in standard form.
2 - 11x² - 8x + 6x²
We can combine like terms, and rewriting it in decreasing power of x's.
⇒ - 11x² + 6x² - 8x + 2
⇒ (-11 + 6)x² - 8x + 2
⇒ -5x² - 8x + 2
Hence, option A is correct i.e. -5x² - 8x + 2; quadratic trinomial.
Solving question 4 :
White-sided jackrabbits: 5.5x² - 9.2x + 6.9
Black-tailed jackrabbits: 5.5x² + 9.9x + 1.3
Total population = White-sided jackrabbits + Black-tailed jackrabbits
Total population = (5.5x² - 9.2x + 6.9 ) + (5.5x² + 9.9x + 1.3)
Total population = (5.5 + 5.5)x² + (9.9 - 9.2)x + (6.9 + 1.3 )
Total population = 11x² + 0.7x + 8.2
Hence, option A is correct i.e. 11x² + 0.7x + 8.2
if 10% of x is 20, what is 23% of x?
Answers
----------
then x = 200.what is 23% of x?
----------
23% = (23/100)*200 = 46
Use common sense to determine whether the given event is impossible; possible, but very unlikely; or possible and likely. A solar eclipse occurs on your birthday.
Answers
-Kaylah
1. Simplify using only positive exponents:
(2t)⁻⁶
2. Simplify using only positive exponents:
(w⁻²j⁻⁴)⁻³(j⁷j³)
3. Simplify using only positive exponents:
a²b⁻⁷c⁴
----------
a⁵b³c⁻²
4. Evaluate the expression for m = 2, t = -3, and z = 0.
z⁻ᵗ(mᵗ)ᶻ
5. Use scientific notation to rewrite the number:
a. 0.0002603 in scientific notation
b. 5.38 × 102 in standard notation
Answers
[tex](2t)^{-c} = \frac{1}{(2t)^{c} } [/tex]
2. To simplify this one, we are going to use the rule for negative exponents twice, the product rule: [tex](a^{n} )(a^{m} )=a^{n+m} [/tex], and the power rule [tex](a^{n}) ^{m} =a^{(n)(m)} [/tex], so:
[tex](w^{-2} j ^{-4} ) ^{-3} (j^{7} j ^{3} )=( \frac{1}{w ^{-2}j ^{-4} } )^{3} (j ^{7+3} )[/tex]
[tex]=(w^2j^4)^3(j ^{10}) [/tex]
[tex]=(w^6j^{12})(j^{10})[/tex]
[tex]=w^6j^{22}[/tex]
3. To simplify this one we are going to use the rule for negative exponents, the product rule, and the quotient rule: [tex] \frac{a^n}{a^m} =a^{n-m}[/tex], so:
[tex] \frac{a^2b^{-7}c^4}{a^5b^3c^-2} = \frac{a^2c^4c^2}{a^5b^3b^7} = \frac{a^{-3}c^6}{b^{10}} = \frac{c^6}{a^3b^10} [/tex]
4. The first thing we need to is apply the exponents rules; in this case our rule for negative exponents:
[tex]z^{-t}(m^t)^z=( \frac{1}{z^t} )(m^{(t)(z)})= \frac{m^{tz}}{z^t} [/tex]
Now can replace our numerical values:
[tex] \frac{2^{(-3)(0)}}{0^{-3}} [/tex]
We have a negative exponent in the denominator, so lets apply oir rule for negative exponents again:
[tex]2^{(-3)(0)}0^3=2^00^3=(1)(0)=0[/tex]
5. Scientific notation is just a way of writing large an small numbers using powers of 10. The exponent of 10 will be the number of places we shift the decimal point to write the number in scientific notation. A positive exponent shows that the decimal point is shifted the right, and a negative one shows that the decimal point is shifted to the left:
a. [tex]0.0002603=2.603[/tex] x [tex]10^{-4} [/tex]
b. [tex]5.38[/tex] x [tex]10^{2} =538[/tex]
Expressions with exponents can be simplified using rules of exponents, and numbers can be converted into scientific notation by recognizing how to move the decimal point and denote magnitude with the power of ten.
To simplify expressions with exponents and convert numbers into scientific notation, we apply the rules of exponents and understand the format of scientific notation.
(2t)⁻⁶: Using the negative exponent rule, which states that a⁻⁶ = 1/a⁶, we can simplify this expression to 1/(2⁶t⁶).
(w⁻²j⁻⁴)⁻³(j⁷j³): To deal with the negative and compounded exponent, we invert and take the cube, resulting in w⁶j¹². Then, multiply the j terms together to get w⁶j¹⁵.
To simplify a²b⁻⁷c⁴ / a⁵b³c⁻², we subtract exponents when dividing like bases, resulting in a⁻³b⁻¹°c⁶.
For the expression z⁻ᵗ(mᵗ)¹, when any variables are raised to the zero power, the result is 1. Thus, the entire expression evaluates to 1 due to (mᵗ)¹ becoming 1.
Converting to scientific notation: To express 0.0002603 in scientific notation, it becomes 2.603 × 10⁻⁴. The number 5.38 × 10² in standard notation is 538.
By applying these step-by-step procedures, we can simplify expressions using positive exponents and accurately convert between standard notation and scientific notation.
A basket contains 4 green marbles and 8 blue marbles. a marble is drawn without replacement. then another marble is drawn. what is the probability that both marbles will be green?
Answers
Final answer:
The probability of drawing two green marbles consecutively without replacement from a basket of 4 green marbles and 8 blue marbles is 0.1, or 10%.
Explanation:
The question involves calculating the probability of drawing two green marbles in succession without replacement from a basket containing 4 green marbles and 8 blue marbles. For the first draw, the probability of drawing a green marble is 4 out of 12, which reduces to 1/3 or about 0.3333. Once that marble is drawn, there are 3 green marbles left and 7 blue marbles, making a total of 10.
Therefore, the probability of drawing another green marble is 3 out of 10, or 0.3. To find the probability of both events happening consecutively, we multiply the two individual probabilities: (1/3) * (3/10) = 1/10 or 0.1. Hence, the probability that both marbles will be green is 0.1, or 10%.
Draw any two convex pentagons. For each of them measure the sum of its interior angles using a protractor. Explain the result of the measuring.
FIRST ANSWER GETS BRAINLIEST ANSWER!
Answers
Answer:
540
Step-by-step explanation:
Milena's take-home pay is $1200 a month. She spends 12% of her take-home pay on her cable bill. How much is Milena's monthly cable bill?
Answers
Answer:
$144
Step-by-step explanation:
Just did test
1.What is the volume of a right circular cylinder with a diameter of 19.6 yd and a height of 23.52 yd?
Enter your answer in the box. Use 3.14 for pi and round only your final answer to the nearest hundredth.
2.What is the volume of a right circular cylinder with a base diameter of 18 yd and a height of 3 yd?
Enter your answer in the box. Express your answer using π .
Answers
The answer should be 7092.822912 or 7092.82 when rounded to the nearest hundredth because the formula for volume is V= π times r^2 times height to get the answer. So: 3.14 x 9.8^2 x 23.52, and that's the answer.
QUESTION 1
We want to find the volume of a circular cylinder with a diameter of [tex]19.6yd[/tex] and a height of [tex]23.52yd[/tex].
The volume of a cylinder is given by the formula
[tex]V=\pi r^2h[/tex]
where [tex]h=23.52yd[/tex] and [tex]r=9.8yd[/tex] is half the diameter of the cylinder and [tex]\pi=3.14[/tex].
We substitute all these values into the formula to obtain,
[tex]V=3.14\times 9.8^2\times 23.52[/tex]
[tex]V=7092.82[/tex] square yards to the nearest hundredth.
QUESTION 2
We want to find the volume of a right circular cylinder with a base diameter of [tex]18yd[/tex] and a height of [tex]3yd[/tex].
The volume of a cylinder is given by the formula
[tex]V=\pi r^2h[/tex]
where [tex]h=3yd[/tex] and [tex]r=9yd[/tex] is half the diameter of the cylinder.
We substitute all these values into the formula to obtain,
[tex]V=\pi \times 9^2\times 3[/tex]
[tex]V=243\pi[/tex] square yards.
Katherine is landscaping her home with juniper trees and pansies. She wants to arrange 15 pansies around each of 8 trees. Each tree costs $20.75 and a six-pack of pansies costs $2.50. Explain how to write an expression to find Katherine’s final cost.
Answers
X-----------------> number of pansies
y-----------------> number of trees
we know that
x=15*8----------> x=120 pansies
y=8 trees
cost of each trees is----------> $20.75
cost of each pansies is------> $2.50/6------> $5/12
[expression to find Katherine’s final cost]=[cost trees]+[cost pansies]
[cost trees]=y*$20.75
[cost pansies]=x*($5/12)
[expression to find Katherine’s final cost]=y*($20.75)+x*($5/12)
[expression to find Katherine’s final cost]=8*($20.75)+120*($5/12)
[expression to find Katherine’s final cost]=$166+$50
[expression to find Katherine’s final cost]=$216
the answer is
[expression to find Katherine’s final cost]=y*($20.75)+x*($5/12)
[expression to find Katherine’s final cost]=8*($20.75)+120*($5/12)
Katherine’s final cost is $216
Answer:
Look below
Step-by-step explanation:
The total cost of the trees must be added to the total cost of the pansies. The tree cost is the cost of one tree times eight. The pansy cost is the cost for 15 pansies multiplied by 8 trees, then divided by the number of pansies in a pack: 20.75(8) + 2.50(15)(8) ÷ 6.
Jerry lost her credit card and instead of reporting it right away, she decides to continue looking for it for a couple of days. On the second day, she makes the call and reports the card lost/stolen to the credit card company. She then logs into the account activity page of his credit card and sees a recent $500 purchase that was made by someone else. How much of this $500 charge will Jerry have to pay?
Answers
The monthly list of expenditures on your credit card statement can be very helpful at tax time to find items for which you are entitled to tax deductions. true or false
Answers
When it comes time to file your taxes, there are a variety of different items that could be deductible.
For example, if you operate your own business, you may be able to deduct certain expense. Also, if you have a lot of medical expenses, they could be tax deductible. If you donate money to charitable organizations, it could be tax deductible.
It would be wise to save your statements and look for deductions.
Answer:
true
Step-by-step explanation:
What is the median for the data set? 252, 210, 264, 278, 208, 295, 248, 257, 284, 271
Answers
208,210,248,252,257,264,271,278,284,295
The median will be the 2 middle numbers because the number of numbers are even.
So the median numbers are 257 and 264.
Hope I helped
I need the answer to question number 12
Answers
.. = (1/3)(32 in)^2*(28 in) ≈ 9557 in^3
1. Which of the following is NOT true about an isosceles trapezoid?
The diagonals are congruent.
The bases are parallel.
The diagonals are perpendicular.
The two non-parallel sides are congruent.
Answers
In isosceles trapezoids, the diagonals are not perpendicular. They are congruent, the bases are parallel, and the non-parallel sides are congruent.
Explanation:An isosceles trapezoid is a type of quadrilateral that has a pair of parallel sides, known as the bases, and the other two sides, not parallel, are of equal length. The statement 'The diagonals are perpendicular' is NOT true for isosceles trapezoids. In isosceles trapezoids, the diagonals are congruent and not perpendicular. Just to put in context, the perpendicular diagonals are a characteristic of rhombuses and not of isosceles trapezoids.
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If x2 - 4 = 45, then x could be equal to
Answers
To find x, we need to simplify.
x^2 -4 = 45
x^2 = 49
sqrt(49) = 7
So, x = 7.
Hope this helps!
9m2-6/5m+c is a perfect square what is the value of c
Answers
The expression given to us is:
[tex] 9m^2-\frac{6}{5}m+c [/tex]
If the above expression is a perfect square then the middle term will have to be 2 times the square root of the first term times the square root of the last term. Thus:
[tex] -\frac{6}{5}m=2\times 3m\times \sqrt{c} [/tex]
[tex] \therefore \sqrt{c}=-\frac{1}{5} [/tex]
Thus, [tex] c=\frac{1}{25} [/tex]
Thus, for 9m^2-(6/5)m+c to be a perfect square, the value of c must be equal to [tex] \frac{1}{25} [/tex] or 1/25
\use the Venn diagram to calculate probabilities.
Which probabilities are correct? Check all that apply.
P(A|C) = 2/3
P(C|B) = 8/27
P(A) = 31/59
P(C) = 3/7
P(B|A) = 13/27
Answers
The two statements that are correct are the first and third options.
In the first choice, we are looking for values that are in A given that they are already a part of B. 14 of the values in B are also in A. This can be reduced to 2/3 as shown.
In the third choice, we are simply looking for the fraction of the entire chart that are in A. There are 31 values in A and 59 in the total chart. Therefore, the fraction 31/59 is correct.
Answer : 1 and 3 are the correct probabilities.
→According to the given Venn diagram.
Total number of elements = 59.
1)P(C)=[tex]\frac{21}{59}[/tex] and [tex]P(A\cap C)=\frac{14}{59}[/tex] then
[tex]P(A|C)=\frac{P(A\cap C)}{P(C)}[/tex][tex]=\frac{\frac{14}{59}}{\frac{21}{59}}=\frac{14}{21}=\frac{2}{3}[/tex]
2)P(B)=[tex]\frac{27}{59}[/tex] and [tex]P(C\cap B)=\frac{11}{59}[/tex] then
[tex]P(C|B)=\frac{P(C\cap B)}{P(B)}[/tex][tex]=\frac{\frac{11}{59}}{\frac{27}{59}}=\frac{11}{27}[/tex][tex]\neq \frac{8}{27}[/tex]
3) P(A) =[tex]\frac{number\ of\ elements\ in\ A}{Total\ elements}=\frac{31}{59}[/tex]
4) P(C) =[tex]\frac{number\ of\ elements\ in\ C}{Total\ elements}=\frac{21}{59}[/tex][tex]\neq \frac{3}{7}[/tex]
5) [tex]P(B|A)=\frac{P(B\cap A)}{P(A)}[/tex][tex]=\frac{\frac{13}{59}}{\frac{31}{59}}=\frac{13}{31}[/tex][tex]\neq \frac{13}{27}[/tex]
Therefore, option 1 and 3 are correct.
How long will it take the ball to reach the ground
Answers
We set the height function equal to zero and solve for t, the time.
h = -16t^2 + 250
-16t^2 + 250 = 0
16t^2 - 250 = 0
[tex] (4t + \sqrt{250})(4t - \sqrt{250}) = 0 [/tex]
[tex] 4t + \sqrt{250} = 0 [/tex] or [tex] 4t - \sqrt{250} = 0 [/tex]
[tex] 4t = -\sqrt{250} [/tex] or [tex] 4t = \sqrt{250} [/tex]
[tex] t = -\dfrac{\sqrt{250}}{4} [/tex] or [tex] t = \dfrac{\sqrt{250}}{4} [/tex]
We discard the negative solution since time must be positive.
The positive solution is
[tex] t = \dfrac{\sqrt{250}}{4} \approx 4.0 [/tex]
Answer: 4.0 seconds
help me please please
Answers
============================================
Explanation:
Point C is the circumcenter of triangle PQR. This means that we can draw a circle centered at C that goes through points P, Q and R at the same time. This circle has a special name: circumcircle.
The segments PC, RC and QC are all radii, so they are the same length. Pick two of the given expressions and set them equal to one another. Then solve for x
I'm going to pick the expressions for PC and RC
PC = RC
3x+7 = 5x-15
5x-3x = 7+15
2x = 22
x = 22/2
x = 11
Answer:
11
Step-by-step explanation:
Which of the following fractions is not in simplest form?
3/4
7/10
9/12
4/15
Answers
it's simplest form is 3/4
First, let's try simplifying all the fractions. . .
3/4: 3 cannot be divided and equal a whole number (only 4 can) This is in simplest form.
7/10: 7 cannot be divided and equal a whole number (only 10 can) this is in simplest form.
(correct answer) 9/12: 9 divided by 3 = 3 and 12 divided by 3 = 4 so this fraction in simplest form is 3/4.
4/15: 15 cannot be divided and equal a whole number (only the 4 can) this fraction is in simplest form.
Final answer:
9/12
Hope this helps you.
Have a great day!
Using the following equation, find the center and radius of the circle by completing the square.
x2 + y2 + 6x − 6y + 2 = 0
center: (−3, 3), r = 4
center: (3, −3) r = 4
center: (3, −3), r = 16
center: (−3, 3), r = 16
Answers
To complete square to a quadratic equation in its standard form we have:
ax2 + bx + c
Completing squares:
P (x) = (x + b / 2) ^ 2 - b ^ 2/4 + c
Keeping this in mind, we can complete square then:
x2 + y2 + 6x - 6y = -2
(x2 + 6x) + (y2 - 6y) = -2
((x + b / 2) ^ 2 - b ^ 2/4 + c) + ((y + b / 2) ^ 2 - b ^ 2/4 + c) = -2
((x + 6/2) ^ 2 - 6 ^ 2/4 + 0) + ((y + (-6) / 2) ^ 2 - (-6) ^ 2/4 + 0) = -2
((x + 3) ^ 2 - 9) + ((y - 3) ^ 2 - 9) = -2
((x + 3) ^ 2) + ((y - 3) ^ 2) - 9 - 9 = -2
((x + 3) ^ 2) + ((y - 3) ^ 2) - 18 = -2
((x + 3) ^ 2) + ((y - 3) ^ 2) = -2 + 18
((x + 3) ^ 2) + ((y - 3) ^ 2) = 16
((x + 3) ^ 2) + ((y - 3) ^ 2) = 4 ^ 2
Answer:
center: (-3, 3), r = 4
Describe the straight line y=9
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Which of the following numbers is not a prime number?
9
3
7
13
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Mark brainliest
Find the hypotenuse of each isosceles right triangle when the legs are of the given measure. 6 sqrt 2
Answers
x = 45°.
There are several ways to go about solving a triangle like this. The best and easiest is simply to memorize that the hypotenuse is exactly root2 times the other sides. Or, each isosceles side is the hypotenuse (c) ÷ root2
[tex]a = b = c \div \sqrt{2} \\ c = a\sqrt{2} \\ c = 6 \sqrt{2} \times \sqrt{2} = 6 \times 2 = 12[/tex]
Another way to do it is the longer proof of Pythagorean Theorem:
[tex] {c}^{2} = {a}^{2} + {b}^{2}... \: \: c = \sqrt{({a}^{2} + {b}^{2})} \\ [/tex]
[tex]c= \sqrt{({6 \sqrt{2}) }^{2} + ({6 \sqrt{2})}^{2}} \\ = \sqrt{(2 \times{(6 \sqrt{2} )}^{2} )} = \sqrt{2(36 \times 2)} \\ c = \sqrt{144} = 12[/tex]
The hypotenuse of the isosceles right triangle is [tex]\( 12 \)[/tex] units.
In an isosceles right triangle, the legs are congruent, and the hypotenuse can be found using the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse [tex]\( c \)[/tex] is equal to the sum of the squares of the lengths of the other two sides [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
Given that each leg of the isosceles right triangle has a measure of [tex]\( 6\sqrt{2} \)[/tex], we can substitute this value into the formula:
[tex]\[ c^2 = (6\sqrt{2})^2 + (6\sqrt{2})^2 \]\[ c^2 = 36 \times 2 + 36 \times 2 \]\[ c^2 = 72 + 72 \]\[ c^2 = 144 \][/tex]
Now, we take the square root of both sides to find the length of the hypotenuse [tex]\( c \):[/tex]
[tex]\[ c = \sqrt{144} \][/tex]
[tex]\[ c = 12 \][/tex]
So, the hypotenuse of the isosceles right triangle is [tex]\( 12 \)[/tex] units.
Gina and Lucy go to the library at 3:30 p.m. They need to be at home at 4:45 p.m. It takes them 15 minutes to walk to the library. How many minutes can they spend at the library?
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Gina and Lucy can spend 45 minutes at the library.
_____
They will get to the library at 3:45. They must start home by 4:30. From 3:45 to 4:00 is 15 minutes, and it is 30 more minutes to 4:30. The time they can spend at the library totals 45 minutes.
what is the midpoint of a segment in the complex plane with endpoints at 6 -2i and -4 + 6i
Answers
Answer:
Midpoint of a segment in the complex plane with endpoints at 6 -2i and -4 + 6i is:
1+2i
Step-by-step explanation:
The midpoint of a segment in the complex plane with endpoints at 6 -2i and -4 + 6i is:
[tex]\dfrac{6-2i-4+6i}{2} \\\\=\dfrac{2+4i}{2}\\ \\=1+2i[/tex]
Hence, midpoint of a segment in the complex plane with endpoints at 6 -2i and -4 + 6i is:
1+2i
Answer: 2 + 4 i
Step-by-step explanation:
Hi, to solve this we have to apply the next expression:
(a1 +a2)/ 2 + (b1 +b2 )/2 i=
Where a is the real part, and b is the imaginary part (with i)
For example, for our case:
6 -2i , 6 is the real part (2) and -2 is the imaginary part (b)
Replacing with the values given
(6 -4) /2+ (-2 +6) /2 i = 2 + 4 i
Feel free to ask for more if needed or if you did not understand something.