Answers
If Disc> 0, the function has two distinct roots
If Disc = 0, the function has a repeated root
If Disc < 0, the function has no real root.
The discriminant can be calculated as:
Disc = b² - 4ac
For the given equation:
Disc = 9 - 4(2)(1) =1
Since the value of discriminant is positive, the function will have two distinct roots.
So, the answer to this question is option B
Related Questions
A 16 ounce package of cereal cost $4.00? what is the unit cost
Answers
Here, we need to find unit cost. Unit cost can be defined as a cost per unit of a product.
We are given that 16 ounce will cost $ 4.00.
That means -
Cost of 16 ounce of cereal package = $ 4.00
Cost of 1 ounce of cereal package -
(We will divide both sides by 16)
Cost of 16 ounce of cereal package ÷ 16 = $ 4.00 ÷ 16
Cost of 1 ounce of cereal package = $ 0.25.
Thus, the cost of 1 ounce of cereal package = $ 0.25.
Answer: 1 oz. Over 0.25 is your answer.
Step-by-step explanation:
the area of a circle is about 254 cm what is the radius
Answers
To find the radius of a circle with an area of 254 cm², you can use the formula for the area of a circle by solving for the radius. We get the radius as 8.99 cm.
The radius of a circle can be found by using the formula for the area of a circle:
Given Area = 254 cm²Formula: A = πr²Solve for r: r = √(A ÷ π)Substitute in Area: r = √(254 cm² ÷ 3.14159)Calculate the radius: r ≈ √(80.865) ≈ 8.99 cmWrite the standard form of the number shown. three hundred eighty-nine billion, nine hundred thirty-six million, two hundred six thousand, seven hundred sixty-seven
Answers
The standard form of the given number is 3.89936206767 × 10¹¹
From the question, the given number is
three hundred eighty-nine billion, nine hundred thirty-six million, two hundred six thousand, seven hundred sixty-seven.
First, we will write the number in figures before writing it in standard form.
The number, three hundred eighty-nine billion, nine hundred thirty-six million, two hundred six thousand, seven hundred sixty-seven, written in figure is
389,936,206,767.
Now, to write in standard form,
A number is said to be in standard form, if it is written in form A × 10ⁿ. Where A is a number bigger than or equal to 1 and less than 10 ( That is, A ≥ 1 and A < 10)
and n is any positive or negative whole number
∴ The number, 389,936,206,767, written in standard form is 3.89936206767 × 10¹¹
Hence, the standard form of the given number is 3.89936206767 × 10¹¹
Learn more here: https://brainly.com/question/1565680
Explain why a counter with an upper limit of five(101) resets at six (110)
Answers
Before going to a higher number than 5 which next is 6(110) or beyond the upper limit, It resets.
Find the equation for the plane through the points upper p 0 left parenthesis negative 2 comma negative 5 comma negative 4 right parenthesis, upper q 0 left parenthesis 5 comma 1 comma negative 4 right parenthesis, and upper r 0 left parenthesis negative 1 comma 2 comma 5 right parenthesis.
Answers
ax +by +cz = 1
By using the given point coordinates for x, y, and z, we end up with three equations in 3 unknowns. These can be described by the augmented matrix ...
[tex] \left[\begin{array}{ccc|c}-2&-5&-4&1\\5&1&-4&1\\-1&2&5&1 \end{array}\right] [/tex]
Putting this in reduced row-echelon form, we find
a = 54/35
b = -63/35
c = 43/35
The equation of the plane through [tex]P_{0}(-2,-5,-4), Q_{0}(5,1,4), R_{0}(-1,2,5)[/tex] is ...
54x -63y +43z = 35
Write 2018 to the power of 2019 + 2018 as the sum of two perfect squares
Answers
According to Brahmagupta-Fibonnaci identity
[tex]( a^{2} + b^{2}) (c^{2}+ d^{2}) = (ac+bd)^{2} + (ad-bc )^{2} [/tex]
Then, we can write that
[tex](13^{2}+ 43^{2})(( 2018^{1009})^{2} + 1^{2})[/tex]
[tex][(13*2008^{1009}+43)^{2} +(13-2008^{1009}*43)^{2}] [/tex]
Q.E.D
Answer: I, III, and IV only
Step-by-step explanation:
How do you solve 6<_2g?
Answers
You're left with 6 / 2 < g which equals 3 < g.
Which set of ordered pairs represents a function?
{(2, –2), (1, 5), (–2, 2), (1, –3), (8, –1)}
{(3, –1), (7, 1), (–6, –1), (9, 1), (2, –1)}
{(6, 8), (5, 2), (–2, –5), (1, –3), (–2, 9)}
{(–3, 1), (6, 3), (–3, 2), (–3, –3), (1, –1)}
Answers
Answer:
{(3, –1), (7, 1), (–6, –1), (9, 1), (2, –1)}
Step-by-step explanation:
A set of ordered pairs in the format [tex](x,y)[/tex] represents a function if for each value of x, there is only one value for y.
The first set of ordered pairs is:
{(2, –2), (1, 5), (–2, 2), (1, –3), (8, –1)}
There are two values of y for [tex]x = 1[/tex]. This means that this set of ordered pairs does not represent a function.
The second set of ordered pairs is:
{(3, –1), (7, 1), (–6, –1), (9, 1), (2, –1)}
For each value of x here, there is only one value of y. This means that this set of ordered pairs represents a function. This is the answer.
The third set of ordered pairs is:
{(6, 8), (5, 2), (–2, –5), (1, –3), (–2, 9)}
There are two values of y for [tex]x = -2[/tex]. This means that this set of ordered pairs does not represent a function.
The fourth set of ordered pairs is:
{(–3, 1), (6, 3), (–3, 2), (–3, –3), (1, –1)}
There are two values of y for [tex]x = -3[/tex]. This means that this set of ordered pairs does not represent a function.
One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number of sterile males that mate with fertile females but produce no offspring. let p represent the number of female insects in a population and s the number of sterile males introduced each generation. let r be the per capita rate of production of females by females, provided their chosen mate is not sterile. then the female population is related to time t by t = p + s p[(r − 1)p − s] dp. suppose an insect population with 10,000 females grows at a rate of r = 1.2 and 400 sterile males are added. evaluate the integral to give an equation relating the female population to time. (note that the resulting equation can't be solved explicitly for p. remember to use absolute values where appropriate.)
Answers
[tex]t = \int { \frac{P+S}{P[(r - 1)P - S]} } \, dP [/tex]
Since r = 1.2 and S = 400, we get:
[tex]t = \int { \frac{P+400}{P[(1.2 - 1)P - 400]} } \, dP [/tex]
= [tex] \int { \frac{P+400}{P(0.2P - 400)} } \, dP [/tex]
This is a rational function that we can write as:
[tex]\frac{P+400}{P(0.2P - 400)} = \frac{A}{P} + \frac{B}{(0.2P - 400)} [/tex]
In order to evaluate A and B, let's calculate the LCD:
[tex]\frac{P+400}{P(0.2P - 400)} = \frac{A(0.2P - 400)}{P(0.2P - 400)} + \frac{BP}{P(0.2P - 400)} [/tex]
which brings to:
P + 400 = 0.2AP - 400A + BP
= P(0.2A + B) - 400A
Since, the left side must be equal to the right side, we get the following system of equations:
[tex] \left \{ {{0.2A + B = 1} \atop {-400A = 400}} \right. [/tex]
Which gives:
[tex] \left \{ {{B=1.2} \atop {A=-1}} \right. [/tex]
Therefore, our integral can be written as:
[tex]t = \int { \frac{P+400}{P(0.2P - 400)} } \, dP = \int {( \frac{-1}{P} + \frac{1.2}{0.2P-400} )} \, dP [/tex]
= [tex]- \int { \frac{1}{P} \, dP + 1.2\int { \frac{1}{0.2P-400} } \, dP[/tex]
= [tex]- \int { \frac{1}{P} \, dP + 6\int { \frac{0.2}{0.2P-400} } \, dP[/tex]
= - ln |P| + 6 ln |0.2P - 400| + C
We now have to evaluate the constant C. In order to do so, we consider that at t = 0 P = 10000:
0 = - ln |10000| + 6 ln |0.2(10000) - 400| + C
C = ln |10000| - 6 ln |1600|
C = ln (10⁴) - 6 ln (2⁶·5²)
C = 4 ln (10) - 36 ln (2) - 12 ln (5)
Therefore, the equation relating female population with time is:
t = - ln |P| + 6 ln |0.2P - 400| + 4 ln (10) - 36 ln (2) - 12 ln (5)
The equation relating females population with time is [tex]\boxed{t = - \ln \left| P \right| + 6\ln \left| {0.2P - 400} \right| - 12\ln \left( 5 \right) + 4\ln \left( {10} \right) - 36\ln \left( 2 \right)}.[/tex]
Further explanation:
Explanation:
The female population is related to time and it is given as follows,
[tex]t = \int {\dfrac{{p + s}}{{p\left[ {\left( {r - 1} \right)p - s} \right]}}dp}[/tex]
The sterile males are 400 and the growth rate is 1.2.
[tex]\begin{aligned}t&= \int {\frac{{p + 400}}{{p\left[ {\left( {1.2 - 1} \right)p - 400} \right]}}dp}\\&= \int {\frac{{p + 400}}{{p\left[ {0.2p - 400} \right]}}dp}\\\end{aligned}[/tex]
The expression [tex]\dfrac{{p + 400}}{{p\left[ {0.2p - 400} \right]}}[/tex] can also be expressed as follows,
[tex]\dfrac{{p + 400}}{{p\left[ {0.2p - 400} \right]}} = \dfrac{A}{p} + \dfrac{B}{{\left( {0.2p - 400} \right)}}[/tex]
Find the least common denominator.
[tex]\dfrac{{p + 400}}{{p\left( {0.2p - 400} \right)}} = \dfrac{{A\left( {0.2p - 400} \right)}}{{p\left( {0.2p - 400} \right)}} + \dfrac{{Bp}}{{p\left( {0.2p - 400} \right)}}[/tex]
[tex]\begin{aligned}p + 400 &= 0.2Ap - 400A + Bp\\p + 400 &= p\left( {0.2A + B} \right) - 400A\\\end{aligned}[/tex]
Equate left and right side of the expression to obtain the value of A and B.
[tex]\begin{aligned}0.2A + B &= 1\\- 400A &= 400\\\end{aligned}[/tex]
The value of A can be obtained as follows,
[tex]\begin{aligned}A&= \dfrac{{400}}{{ - 400}}\\A& =- 1\\\end{aligned}[/tex]
The value of B is 1.2.
Integral can be expressed as follows,
[tex]\begin{aligned}t &= \int {\frac{{p + 400}}{{p\left( {0.2p - 400} \right)}}dp} \\ &= \int {\left( {\frac{{ - 1}}{p} + \frac{{1.2}}{{\left( {0.2p - 400} \right)}}} \right)dp} \\&= - \int {\frac{1}{p}dp} + 6 \times \int {\frac{{0.2}}{{0.2p - 400}}dp}\\&= - \ln \left| p \right| + 6\ln \left| {0.2p - 400} \right| + C\\\end{aligned}[/tex]
At initial time the population is 10000.
[tex]\begin{aligned}- \ln \left| {10000} \right| + 6\ln \left| {0.2 \times 10000 - 400} \right| + C &= 0\\\ln \left| {10000} \right| - 6\ln \left| {1600} \right| &= C\\4\ln 10 - 36\ln \left( 2 \right) - 12\ln \left( 5 \right) &= C\\\end{aligned}[/tex]
Substitute [tex]4\ln 10 - 36\ln \left( 2 \right) - 12\ln \left( 5 \right)[/tex] for [tex]C.[/tex]
[tex]t = - \ln \left| p \right| + 6\ln \left| {0.2p - 400} \right| + 4\ln 10 - 36\ln \left( 2 \right) - 12\ln \left( 5 \right)[/tex]
The equation relating females population with time is [tex]\boxed{t = - \ln \left| P \right| + 6\ln \left| {0.2P - 400} \right| - 12\ln \left( 5 \right) + 4\ln \left( {10} \right) - 36\ln \left( 2 \right)}.[/tex]
Learn more:
Learn more about inverse of the functionhttps://brainly.com/question/1632445. Learn more about equation of circle brainly.com/question/1506955. Learn more about range and domain of the function https://brainly.com/question/3412497
Answer details:
Grade: High School
Subject: Mathematics
Chapter: Exponential function
Keywords: exponential growth, growth rate model, pesticides, slowing, insect population, sterile males, mate, fertile, females, explicitly, female grows, integral, female insects, male insects, capita rate, chosen mate, generation.
Use any method to solve the equation. If necessary, round to the nearest hundredth.
x^2 + x − 30 = 0
A. –5, 6
B. 10, –12
C. 5, –6
D. 5. 5, –5.5
Answers
Here, in order to find the solutions to the quadratic equation - x² + x - 30 = 0, we will use factorization method.
In the method, we will split 30, in such factors, which when added or subtracted gives us 1, and when multiplied gives us -30.
So, we will use, -5 and 6. When they are added they will give us 1 and when multiplied they will give us -30 as answer.
Now, the equation will be written as -
x² - 5x + 6x - 30 = 0
Taking common, we get
x(x - 5) +6(x-5) = 0
(x-5)(x+6) = 0
So, x - 5 = 0 and x +6 = 0, we will get, x = 5 and x = - 6
Thus, the correct option is C). x = 5 and -6
Find the total cost of a $619.50 refrigerator if the sales tax rate on the purchase is 2.8?
Answers
What is the solution of square root of-4x=100? x = –2500 x = –50 x = –2.5 no solution
Answers
-4x = 100² = 10,000
x = 10,000/-4 = -2500
The solution is ...
x = -2500
the floor sheerly bedroom is a square with a perimeter of 60 ft what is the area of the floor of sherry bedroom
Answers
Write the equation of a circle with center upper c left-parenthesis 4 comma negative 5 right-parenthesis that passes through upper p left-parenthesis 1 comma negative 2 right-parenthesis number .
Answers
(4-1)² +(-5-(-2))² = 18
This is the square of the radius of the circle, so we can use it for r² in the formula
(x -h)² +(y -k)² = r² . . . . . . . circle of radius r and center (h, k)
Your circle has equation ...
(x -4)² +(y +5)² = 18
# 11 Q . please solve the diagrama
Answers
The first choice is the correct
The area of a circle is = π r²
Substitute the given,which will give you this : π * 18² = 324 π in²
You plant a tree that is 36 inches tall. After one year, the tree is 43 inches tall. Which expression describes the percent of increase in the tree's height?
Answers
36 - 100%43 - x %
we multiply by cross -> 36 * x = 100 / 43
we should remove '100%' from new value, because we find only increase.. no new value in percent.
x + 100 = (43 * 100)/36
x = 4300/36 - 100x = 119+ 16/36 -100x = 19 +4/9
the answer is
19.44% increase
Percent Problem:
You need to calculate percent increase from 36 to 43.
First Step: Find the difference between two numbers, in this case, it's 10 - 2 = 8.
Second Step: Take the difference, 7, and divide by the original number: 7/36 = 0.194444.
Last, convert the decimal to a percent.
FINAL ANSWER: 19.44%
Q9 Q13.) Find the products AB and BA to determine whether B is the multiplicative inverse of A.
Answers
See pictures below for the answer!
P.S. I think I answered this question already. I still have it on Microsoft Word :)
As always, it is my pleasure to help students like you!
Brian wants to fence in his triangular plot of farm land that measures 1.1 by 1.5 by 2.2 miles. Determine the angles at which the fences of the three sides will meet.
Rounding each angle to the nearest degree: m<A=___degrees
Answers
The angle measures are ...
∠A = 115°
∠B = 27°
∠C = 38°
_____
You can use the Law of Cosines to solve for angle A, then the Law of Sines for the others.
A = arccos((1.5²+1.1²-2.2²)/(2*1.1*1.5)) = arccos(-1.38/3.3) ≈ 115°
C = arcsin(1.5/2.2*sin(115°)) ≈ 38°
B = 180° -115° -38° = 27°
The angles for Brian's triangular plot of farmland are approximately 27° for ∠A, 38° for ∠B, and 115° for ∠C.
To determine the angles of a triangular plot of land with sides measuring 1.1 miles, 1.5 miles, and 2.2 miles, we can use the Law of Cosines. The Law of Cosines states:
⇒ c² = a² + b² - 2ab × cos(C)
Let's label the sides as follows:⇒ a = 1.1 miles
⇒ b = 1.5 miles
⇒ c = 2.2 miles
First, we calculate ∠C (opposite side c):⇒ cos(C) = (a² + b² - c²) ÷ (2ab)
⇒ cos(C) = (1.1² + 1.5² - 2.2²) ÷ (2 × 1.1 × 1.5)
= (1.21 + 2.25 - 4.84) ÷ (3.3)
= (-1.38) ÷ (3.3)
= -0.4182
Then, we find ∠C by taking the inverse cosine (cos-1):⇒ ∠C ≈ 115°
Next, we calculate ∠A (opposite side a) using the same method:⇒ cos(A) = (b² + c² - a²) ÷ (2bc)
⇒ cos(A) = (1.52 + 2.22 - 1.12) ÷ (2 × 1.5 × 2.2)
= (2.25 + 4.84 - 1.21) ÷ (6.6)
= (5.88) ÷ (6.6)
= 0.8909
Then, we find angle A:⇒ ∠A ≈ 27°
Finally, we calculate angle B (opposite side b) using:⇒ ∠B = 180° - ∠A - ∠C
= 180° - 27° - 115°
= 38°
The angles at which the fences of the three sides will meet are approximately 27° for ∠A, 38° for ∠B, and 115° for ∠C, when rounded to the nearest degree.
Complete question:
Brian wants to fence in his triangular plot of farm land that measures 1.1 × 1.5 × 2.2 miles.
Determine the angles at which the fences of the three sides will meet.
Rounding each angle to the nearest degree:
write percent as a fraction or in simplest form of 37.5%
Answers
37.5% = 37.5/100 = 375/1000 = 3/8
The simplest form is 3/8, therefore the answer to your question is 3/8.
0.375 = 37.5%
Answer = 3/8
~Hope this helped~
Q2 Q13.) Use the given conditions to write an equation for the line in point-slope form and slope-intercept form.
Answers
A (-5,-4) and B (5,10)
Part 1) write an equation for the line in point-slope form
we know that
the equation of a line in point-slope form is
y-y1=m*(x-x1)
step 1
find the slope m
m=(y2-y1)/(x2-x1)------> m=(10+4)/(5+5)----> m=14/10---> 7/5
strep 2
with m=7/5 and the point B (5,10)
find the equation of a line
y-y1=m*(x-x1)-----> y-10=(7/5)*(x-5)
the answer Part 1) is
y-10=(7/5)*(x-5)
Part 2) write an equation for the line in slope-intercept form
we know that
the equation of a line in slope-intercept form is
y=mx+b
step 1
find the slope m
m=(y2-y1)/(x2-x1)------> m=(10+4)/(5+5)----> m=14/10---> 7/5
step 2
with m=7/5 and the point B (5,10)
find b
y=mx+b
10=(7/5)*5+b-----> 10=7+b-------> b=3
the equation of the line is
y=(7/5)x+3
the answer Part 2) is
y=(7/5)x+3
a student uses a solution that contains 16 grams of water to conduct an evaporation experiment
Answers
the correct option is:
b.
Step 1:0.035 xx16=0.56
Step 2: 16-0.56=15.44
Step 3: 0.0425 xx15.44=0.6562
Step 4: 15.44-0.6562=14.7838
Let's break down the calculations:
Step 1: Calculate the amount of water lost at the end of the first hour.
[tex]\[ \text{Amount lost} = 0.035 \times 16 = 0.56 \][/tex]
Step 2: Subtract the amount lost from the initial amount to find the remaining water at the end of the first hour.
[tex]\[ \text{Remaining water} = 16 - 0.56 = 15.44 \][/tex]
Step 3: Calculate the amount of water lost at the end of the second hour.
[tex]\[ \text{Amount lost} = 0.0425 \times 15.44 = 0.6562 \][/tex]
Step 4: Subtract the amount lost from the remaining water at the end of the first hour to find the remaining water at the end of the second hour.
[tex]\[ \text{Remaining water} = 15.44 - 0.6562 = 14.7838 \][/tex]
So, the correct option is:
b.
Step 1:0.035 xx16=0.56
Step 2: 16-0.56=15.44
Step 3: 0.0425 xx15.44=0.6562
Step 4: 15.44-0.6562=14.7838
The complete Question is given below:
A student uses a solution that contains 16 grams of water to conduct an evaporation experiment.
At the end of one hour, the amount of water in the solution has decreased by 3.5% .
At the end of two hours, the amount of water in the solution has decreased by another 4.25% .
Which calculations can be used to determine the amount of water, in grams, remaining in the solution at the end of the second hour?
a.
Step 1:0.035 xx16=0.56
Step 2: 16-0.56=15.44
Step 3: 0.0425 xx15.44=0.6562
Step 4:16-0.6562=15.3438
b.
Step 1:0.035 xx16=0.56
Step 2: 16-0.56=15.44
Step 3: 0.0425 xx15.44=0.6562
Step 4: 15.44-0.6562=14.7838
c.
Step 1:0.35 xx16=5.6
Step 2: 16-5.6=10.4
Step 3: 0.425 xx10.4=4.42
Step 4:16-4.42=11.58
d.
Step 1: 0.35 xx16=5.6
Step 2: 16-5.6=10.4
Step 3: 0.425 xx10.4=4.42
Step 4:10.4-4.42=5.98
A school typically sells 500 yearbooks each year for 50 dollars each. The economic calls does a project and discovers that they can sell 100 more yearbooks for every $5 decrease in price. The revenue for yearbook sales is equal to the number of yearbooks sold times the price of the yearbook. Let x represent the number of $5 decrease in price. If the expression that represents the revenue is written in the form R(x)=(500+ax)(50-bx). Find the value of a and b
Answers
Final answer:
For the expression R(x)=(500+ax)(50-bx), a represents the increase in yearbooks sold for each $5 decrease in price and b represents the corresponding decrease in price per yearbook. The correct values for a and b are 100 and 5 respectively, resulting in the revenue expression R(x) = (500 + 100x)(50 - 5x).
Explanation:
The student's question revolves around the concept of revenue generation based on the number of items sold and the selling price per item.
When examining revenue, we understand it as the product of price per unit times the number of units sold. In the given example, a school is selling yearbooks with a base scenario of 500 yearbooks at $50 each. We are given that for every $5 decrease in price, the school can sell an additional 100 yearbooks. This can be formulated as:
R(x) = (500 + ax)(50 - bx)
Here x represents the number of $5 decreases from the original price. With each decrease, the number of books sold increases by 100, and the price decreases by $5.
Therefore, variable a must represent the increase in quantity sold for each $5 price decrease (a = 100), and variable b must represent the decrease in price per yearbook for each price decrease (b = 5).
Therefore, the values of a and b are 100 and 5, respectively.
The expression for the revenue becomes:
R(x) = (500 + 100x)(50 - 5x)
Q # 4 Solve the graph please <6. , <1. , <2. ,<4
Answers
∠7 + ∠4 = 180°
HELP ASAP PLEASE
Name the quadrant in which the point (x, y) lies.
x > 0 and y < 0
III
II
IV
I
Answers
y < 0 refers to the bottom half of the coordinate plane.
Both these conditions are met in the lower right quadrant of the coordinate plane. Quadrants are numbered counterclockwise from upper right, so the lower right quadrant is number IV.
The name of the quadrant in which (x >0, y < 0) lies is
IV
Answer: quadrant 4
Step-by-step explanation:
Jane multiplied 825 x 22 and got 3,300. Flynn multiplied the same numbers and got 18,150. Which student is correct? What mistake did the other student make?
Answers
Please help me with this
Answers
The explanation is shown below:
1. By definition, a geometric sequence is a sequence of numbers each number, after the first one, is the result of multiply the previous number by a common ratio.
2. The Option B has the following common ratio:
(2/3) / (4/9)=1.5
1/(2/3)=1.5
r=1.5
3. The Option E has the following common ratio:
-15/3=-5
75/-15=-5
r=-5
Hey can you please help me posted picture of question
Answers
So, 2, 2, 2 = B
See attached picture:
Any and all answers get a 5 star rating
First answer gets a thanks
Most reliable answer gets brainliest
The following data shows the weight, in pounds, of 6 boxes:
5, 3, 3, 4, 5, 4
What is the value of the mean absolute deviation of the weight of the boxes, and what does it represent about the weight of a box?
Answers
Hope this Helps!!
Density is mass divided by volume. Find the mass of a gold bar if the density is 19.32 g/cm3 and the volume is 50 cm3.
Answers
mass = density * volume
mass = 19.32 * 50
mass = 966 grams
Answer:
d
Step-by-step explanation:
denisity * volume = mass
In a right rectangles pyramid with base edges a= 18 cm and b= 10 cm the slant height toward a is k= 13 cm while the slant height towards b is l= 15 cm. What is the surface area of the pyramid
Answers
rectangular base:
A = lw
18 x 10
A = 180 square cm
side a triangular lateral faces:
A = 1/2bh
1/2 x 18 x 13
A = 117 square cm x 2= 234 square cm
side b triangular lateral faces:
A = 1/2bh
1/2 x 10 x 15
A = 75 square cm x 2= 150 square cm
Add all the bold areas together:
150 + 234 +180 = 564
The surface area is 564 square cm.