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3 years ago

I am having some trouble with finding out the two problems for this question, can you please help me?

Mathematics

1 answer:

sattari [20]3 years ago

4 0

What is it telling us to do- and what do we know?

To find the perimeter of the problem, add all side lengths together.

To find the area, add all the areas of the piece together.

Imagine your taking the piece apart.

Find the area of each separate piece.

to find the width of c and d, divide 70 by 14. That = 5

There are five separate widths-b- c- d and all in between, so we can safely assume each width is 14.

To find the area of b, multiply 50-12 by 14. 38*14=542

To find the area of c, multiply 25 by 14=350

To find the area of d, multiply 50 by 14, because at the top we can see its equal to 50. That =700

Then finally, to find the area of a, multiply 70 by 12 to get 840. Added all together, the total area is 2432.

To find the perimeter, we take all the data we've gathered about the side lengths and add it together. The total perimeter is 414.

Hope this helps!

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Evaluate the expression 8 power of 2+3.8-10 times 2

Nitella [24]

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

-----------------------------------------------------------------------

$8^2+3.8-10\cdot2=64+3.8-20=67.8-20=47.8$

6 0

3 years ago

What are the restrictions for a? 2a^2+a-15/5a^2+16a+3

koban [17]

ANSWER

The restrictions are

$a\ne -3,a\ne -\frac{1}{5}$

EXPLANATION

We were given the rational function,

$\frac{2a^2+a-15}{5a^2+16a+3}$

The function is defined for all values of a, except

$5a^2+16a+3=0$

This has become a quadratic trinomial, so we need to split the middle term.

We do that by multiplying the coefficient of $x^2$ which is 5 by the constant term which is 3. This gives us 15.

The factors of 15 that adds up to 16 are 1 and 15.

We use these factors to split the middle term.

$5a^2+15a+a+3=0$

We now factor to get,

$5a(a+3)+1(a+3)=0$

We factor further to get,

$(a+3)(5a+1)=0$

This implies that,

$(a+3)=0,(5a+1)=0$

This gives

$a=-3,a=-\frac{1}{5}$

These are the restrictions.

8 0

3 years ago

Calculate the average rate of change of the function f(x) = 3x2 over the interval 1 ≤ x ≤ 5.

RideAnS [48]

Answer:

Step-by-step explanation:

The average rate of change of f(x) in the closed interval [ a, b ] is

$\frac{f(b)-f(a)}{b-a}$

here [ a, b ] = [ 1, 5 ]

f(b) = f(5) = 3 × 5² = 75

f(a) = f(1) = 3 × 1² = 3, hence

average rate of change = $\frac{75-3}{5-1}$ = $\frac{72}{4}$ = 18

5 0

3 years ago

Find the coordinates of the other endpoint of AB with A(6,-4) and (3,10) as the midpoint

Oxana [17]

Answer is B(0,24) . I try my best to help you . Hope can help you.

3 0

3 years ago

12What mistake did the student make when solvingtheir two-step equation?(a)b) If correctly solved what should the value of be?

mixas84 [53]

Given the equation:

$\frac{x}{6}+3=-18$

(a) You can identify that the student applied the Subtraction Property of Equality by subtraction 3 from both sides of the equation:

$\frac{x}{6}+3-(3)=-18-(3)$

However, the student made a mistake when adding the numbers on the right side.

Since you have two numbers with the same sign on the right side of the equation, you must add them, not subtract them and use the same sign in the result. Then, the steps to add them are:

- Add their Absolute values (their values without the negative sign).

- Write the sum with the negative sign.

Then:

$\frac{x}{6}=-21$

(b) The correct procedure is:

1. Apply the Subtraction Property of Equality by subtracting 3 from both sides (as you did in the previous part):

$\begin{gathered} \frac{x}{6}+3-(3)=-18-(3) \\ \\ \frac{x}{6}=-21 \end{gathered}$

2. Apply the Multiplication Property of Equality by multiplying both sides of the equation by 6:

$\begin{gathered} (6)(\frac{x}{6})=(-21)(6) \\ \\ x=-126 \end{gathered}$

Hence, the answers are:

(a) The student made a mistake by adding the numbers -18 and -3:

$-18-3=-15\text{ (False)}$

(b) The value of "x" should be:

$x=-126$

4 0

1 year ago