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1 year ago

14

Margo's parents started a savings account when she started highschool. They put in $12,500 to start. If the account earns 7% sim

ple interest every year, How much will she have after 4 years?

1 answer:

eduard1 year ago

7 0

Answer:

probably bout like 4 dollars

Step-by-step explanation:

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a construction worker has rope 9 m long he needs to cut it in pieces that are each 3 /5 m long how many such pieces can he cut w

lapo4ka [179]

Dividing a whole number and an improper fraction seems tricky, but it just has one or two more steps than the normal process. First, convert the whole number "9" into a fraction by giving it a denominator of "1". It should look like 9/1. Now that both numbers are converted into like formats (improper fractions), you should have 9/1 divided by 5/3.

Second, flip 5/3 over into 3/5 - this is the "reciprocal". Now you have 9/1 divided by 3/5. It's just a matter of multiplying across the math sentence. Multiply the numerators (9x3) and the denominators (1 x 5). Your new fraction should be 27/5. This is your answer in the improper fraction format.

You can create other formats depending on the expected answer. For a mixed number, "divide your fraction UP" (27 divided by 5) which gives you 5 and 2/5. This can be further converted into 5.4 if you need your answer in decimal form.

4 0

3 years ago

Vance is designing a garden in the shape of an isosceles triangle. The base of the garden is 30 feet long. The function y = 15 t

Stella [2.4K]

Answer:

A. 8.66 feet

B. 12.59 feet

C. Area of triangle when $\theta=30$ is 129.9 square feet. Area of triangle when $\theta=40$ is 188.85 square feet. Increasing the angle $\theta$ increases the area.

Step-by-step explanation:

The equation that models the height of the triangle is:

$y=15 Tan \theta$

Where,

$y$ is the height, and

$\theta$ is the angle

A.

When $\theta=30$ , the height is:

$y=15Tan30\\y=8.66$

B. When $\theta=40[/tex\ , the height is:[tex]y=15Tan40\\y=12.59$

C. <em>To find the area of the isosceles triangular shaped garden, we use the </em><em>formula for the area of the triangle</em><em>:</em>

$A=\frac{1}{2}bh$

Where,

A is the area
b is the base, which is given as 30 feet, and
h is the height [8.66 feet when the angle is 30 & 12.59 when angle is 40]

<u>When Vance uses $\theta=30$ , the area is</u>:

$A=\frac{1}{2}(30)(8.66)\\A=129.9$ square feet

<u>When Vance uses $\theta=40$ , the area is</u>:

$A=\frac{1}{2}(30)(12.59)\\A=188.85$ square feet

So we see that when the angle is more, the area is also more.

4 0

3 years ago

Read 2 more answers

11111111111111111111111

Nat2105 [25]

Answer:

1) Zero based on (-16·t - 2) is t = -1/8 second

2) Zero based on (t - 1) is t = 1 second

Step-by-step explanation:

The given functions representing the height of the beach ball the child throws as a function of time are;

y = (-16·t - 2)·(t - 1) and y = -16·t² + 14·t + 2

We note that (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

Therefore, the function representing the height of the beachball, 'y', is y = (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

The zeros of a function are the values of the variables, 'x', of the function that makes the value of the function, f(x), equal to zero

In the function of the question, we have;

y = (-16·t - 2)·(t - 1) = -16·t² + 14·t + 2

The above equation can be written as follows;

y = (-16·t - 2) × (t - 1)

Therefore, 'y' equals zero when either (-16·t - 2) = 0 or (t - 1) = 0

1) The zero based on (-16·t - 2) = 0, is given as follows;

(-16·t - 2) = 0

∴ t = 2/(-16) = -1/8

t = -1/8 second

The zero based on (-16·t - 2) is t = -1/8 second

2) The zero based on (t - 1) = 0, is given as follows;

(t - 1) = 0

∴ t = 1 second

The zero based on (t - 1) is t = 1 second

4 0

2 years ago

Which inequality is used when the situation involves a "minimum".

Kay [80]

If the value, let's call it x, is less than a value, it is x<__ if it's less than or equal to a value, use <u><</u>

3 0

2 years ago

If f(x) = -2x + 8 and g(x) = x + 7, what is<br> F•g)(2)?

Blababa [14]

16•7•2 is not the answer

5 0

2 years ago