Write a proportion to find how many points xx a student needs to score on a test worth 80 points to get a test score of 80%.

Answer: 70 I just took the test!

Step-by-step explanation:

Answer:

C. [tex]\dfrac{1}{(m-4)(m-3)}[/tex]

Step-by-step explanation:

First note that

[tex]m^2-16=(m+4)(m-4) \quad \text {and} \quad m^2-9=(m+3)(m-3)[/tex]

On the other hand, the expression from the question is equivalent to

[tex]\dfrac{m+3}{m^2-16} \cdot \dfrac{m+4}{m^2-9}=\dfrac{m+3}{(m+4)(m-4)}\dfrac{m+4}{(m+3)(m-3)}[/tex]

we cancel the equal terms in the numerator and the denominator and finally we get that our expression is equivalent to

[tex]\dfrac{1}{(m-4)(m-3)}[/tex]

Answer:

  $5308.79

Step-by-step explanation:

The future value can be computed from ...

  FV = P(1 +r/n)^(nt)

where P is the principal invested, r is the annual interest rate, n is the number of times per year it is compounded, and t is the number of years.

You have P = $5000, r = 0.03, n = 12 (months per year), t = 2.

Filling in the given numbers, we have ...

  FV = $5000(1 +.03/12)^(12·2) ≈ $5000(1.0617570) ≈ $5308.79

The amount of the withdrawal will be $5308.79.

Final answer:

To find the withdrawal amount for Myron's CD, the compound interest formula is used with the respective values plugged in. After 2 years, with 3% interest compounded monthly on a principal of $5000, the withdrawal amount is approximately $5,304.91.

Explanation:

The question deals with the calculation of compound interest for a Certificate of Deposit (CD), specifically with different compounding periods.

To calculate the withdrawal amount for Myron's 2-year CD with a principal amount of $5000 at a 3% interest rate compounded monthly, the formula for compound interest is:

[tex]A = P (1 + r/n)^{nt[/tex]

Where:

By plugging in Myron's values, we get:

[tex]A = $5000 (1 + 0.03/12)^{12\times 2[/tex]

Calculating this we find that A, the withdrawal amount, is approximately $5,304.91.

The two positive consecutive even integers that, when their squares are subtracted, equal 76 are 18 and 20.

In this problem, the two positive consecutive even integers can be represented as x and x+2. The problem states that the difference of their squares is 76, which can be put into the equation (x + 2)^2 - x^2 = 76. By simplifying this equation, we get 4x + 4 = 76. Solving for x, we find that x = 18. Therefore, the two consecutive even integers are 18 and 20.

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o to 100 miles per hour

1(17/24) simplified is 41/24.

We have,

To simplify the mixed number 1 17/24, we need to convert it to an improper fraction.

To do that, we multiply the whole number (1) by the denominator of the fraction (24), and then add the numerator (17):

1 x 24 + 17 = 24 + 17 = 41

So, the mixed number 1 17/24 is equivalent to the improper fraction 41/24.

Therefore,

1(17/24) simplified is 41/24.

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The assertion is not totally true.

It is not always necessary to dispose of debris dropped more than 20 feet to the outside using an enclosed chute.

It is crucial to check the precise restrictions that apply in your area as building codes and regulations might differ based on the jurisdiction.

It is important to keep in mind that, in some circumstances, a covered chute may be necessary for safety reasons or to stop the materials from spreading or causing harm while being disposed of.

These specifications may be unique to certain locales, building types, or construction projects.

Therefore, it is crucial to check your area's construction laws and regulations to ascertain the precise specifications for disposal requirements.

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Answer:

Step-by-step explanation:

2^3)(2^-4) = 2^(3-4) = 2^-1 = 1/2

answer is D.

add exponents and keep the same base.then find the reciprocal and change the sign of the exponent

Answer:

Vertical angles : Opposite angles made by two intersecting lines.

Corresponding angles : which occupy the same relative position at each intersection where a straight line crosses two others.

Alternate interior angles : when a transversal passes through two line then the angles that are formed on opposite sides of the transversal and inside the two lines.

Alternate exterior angles : angles on the outer side of each of those two lines but on opposite sides of the transversal.

Here, line k passes through the lines p and q,

Thus, by the above explanation,

Vertical pairs are, angle 1 and angle 4, angle 2 and 3, angle 5 and angle 8,  angle 6 and 7,

Corresponding pairs,

angle 2 and angle 6, angle 1 and angle 5, angle 4 and angle 8, angle 3 and angle 7,

Alternate interior angles pair,

angle 4 and angle 5, angle 6 and angle 3,

Alternate exterior angles pair,

angle 2 and angle 8, angle 1 and angle 7,

Vertical angles are always equal in measure,

Now, the p and q are parallel lines,

So, the corresponding angles will be congruent as well as alternative interior or exterior will be congruent.

Therefore, the required pair,

Corresponding pairs :

m∠1=m∠5

m∠2=m∠6

Alternative interior pairs :

m∠3=m∠6

m∠4=m∠5,

Alternate exterior pairs :

m∠1=m∠8

m∠2=m∠7

Vertical pairs :

m∠1=m∠4

m∠5=m∠8

m∠2=m∠3

m∠6=m∠7

Answer:

its reflection

Step-by-step explanation:

I did it on edge ; )

Answer:

9,800

Step-by-step explanation:

Helen's Shop

60 m x 80 m = 4,800 m2

Parking Lot

Length = 2 × 50 m = 100 m

Area = 50 m × 100 m = 5,000 m2

Final Answer

4,800 m2 + 5,000 m2 = 9,800 m2

The rule which describes the translation of the figure is T–3, –2(x, y).

The translation T(x, y) → (x – h, y – k) moves a point (x, y) to a new point (x – h, y – k).

T–3, –2(x, y) moves the point 3 units to the right and 2 units down.

T–3, 2(x, y) moves the point 3 units to the right and 2 units up.

T3, –2(x, y) moves the point 3 units to the left and 2 units down.

T3, 2(x, y) moves the point 3 units to the left and 2 units up.

So the answer is T–3, –2(x, y).

Here is a table that shows the effect of each translation on the point (3, 4):

Original point | Translation | New point

(3, 4) | T–3, –2(x, y) | (0, 2)

(3, 4) | T–3, 2(x, y) | (6, 6)

(3, 4) | T3, –2(x, y) | (6, 2)

(3, 4) | T3, 2(x, y) | (0, 6)

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The hang out time is 0.875 sec.

An algebraic equation of the second degree in x is a quadratic equation. The quadratic equation is written as ax² + bx + c = 0, where x is the variable, a and b are the coefficients, and c is the constant term. First, there must be a term other than zero in the coefficient of x²(a>0) for an equation to be a quadratic equation. The x² term is written first when constructing a quadratic equation in standard form, then the x term, and finally the constant term. The numerical values of a, b, and c are often expressed as integral values rather than fractions or decimals.

Given Equation:

h(t) = –16t² + 14t

For the Hangout time we can consider the function h(t)=0

0 =  –16t² + 14t

16t² = 14t

16t= 14

t= 14/16

t= 0.875 sec

Hence, the hang out time is 0.875 sec.

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A student concluded that the solution of the equation [tex]\sqrt{2x+1}+3 = 0[/tex] is x = 4. The conclusion is true.

Perfect squares are those integers whose square root is an integer.

Let x-a be the closest perfect square less than x,

and let x+b be the closest perfect square more than x, then we get x-a < x < x+b (no perfect square in between x-a and x+b, except possibly x itself).

Then, we get:

[tex]\sqrt{x-a} < \sqrt{x} < \sqrt{x+b}[/tex]

[tex]\sqrt{2x+1}+3 = 0\\\\\sqrt{2x+1}= -3[/tex]

by taking square on both sides we get

[tex](\sqrt{2x+1})^2 = (- 3)^2 \\\\2x+ 1 = 9[/tex]

when x = 4

2(4) + 1

= 9

Thus, the solution of the equation is 4.

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