Math please help what is the area

Use greatest common factor and the distributive property to write equivalent expressions in factored form for the following expr

Nitella [24]

Answer:

a. 4(d+3e)

b. 6(3x+5y)

c. 7(3a+4y)

d. 8(3f+7g)

Step-by-step explanation:

In each case we find the greatest common factor of the numbers. That is the greatest number that goes into both the numbers. Then we factor it out in front and inside parentheses we divide each original term by the greatest common factor:

a. 4d+12e, GCF: 4

$\displaystyle4\left(\frac{4d}{4}+\frac{12e}{4}\right)=4(d+3e)$

b. 18x+30y, GCF: 6

$\displaystyle6\left(\frac{18x}{6}+\frac{30y}{6}\right)=6(3x+5y)$

c. 21a+28y, GCF: 7

$\displaystyle7\left(\frac{21a}{7}+\frac{28y}{7}\right)=7(3a+4y)$

d. 24f+56g, GCF: 8

$\displaystyle8\left(\frac{24f}{8}+\frac{56g}{8}\right)=8(3f+7g)$

5 0

4 years ago

Which of these triangles appear not to be congruent to any others shown here ? Check all that apply.

Ymorist [56]

The answer is B.
*Hope this helps!*

4 0

4 years ago

Read 2 more answers

Help me pleaseeee.........

ruslelena [56]

9514 1404 393

Answer:

y = 20
4^8
x = 0

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

a^0 = 1 . . . . for a ≠ 0

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And, it is convenient to know the cubes of small integers:

1³ = 1; 2³ = 8; 3³ = 27; 4³ = 64; 5³ = 125

6³ = 216; 7³ = 343; 8³ = 512; 9³ = 729; 10³ = 1000

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1) p^3 × p^5 = p^-12 × p^y

Equating exponents:

3 + 5 = -12 + y

20 = y . . . . . . . . . . . add 12

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2) 64 × 4^5 = 4^3 × 4^5 = 4^(3+5) = 4^8

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3) 10^x = 1 = 10^0

x = 0

6 0

3 years ago

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Enter the letters of the points that lie on the line $y = \frac{1}{2} x + 3.$[asy]unitsize(0.4 cm);pair A, B, C, D, E, F, G, H,

igor_vitrenko [27]

The points that lie on the line are C(-4, 1), E(-6, 6) and I(2, 4)

Given the equation of a line expressed as $y=\frac{1}{2}x + 3$

We need to find which of the points from A to J lies on the line.

To do that, we will substitute the x-coordinate point into the equation to see if we are going to get the y-coordinate

For the coordinate C(-4, 1)

when x = -4

y = 1/2(-4) + 3

y = -2 + 3

y = 1

This shows that the point C(-4, 1) is on the line

For the point E(6, 6)

when x = 6

y = 1/2(6) + 3

y = 3 + 3

y = 6

This shows that the point E(-6, 6) is on the line

For the point I (2, 4)

when x = 2

y = 1/2(2) + 3

y = 1 + 3

y = 4

This shows that the point I(2, 4) is on the line

This shows that the points that lie on the line are C(-4, 1), E(-6, 6) and I(2, 4)

Learn more here: brainly.com/question/16131866

3 0

3 years ago

Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.7

Zinaida [17]

Answer:

(a) The probability that the sample average sediment density is at most 3.00 is 0.092.

The probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b) The sample size must be at least 77.

Step-by-step explanation:

The random variable X ca be defined as the sediment density (g/cm) of a specimen from a certain region.

The random variable X is normally distributed with mean, μ = 2.7 and standard deviation, σ = 0.75.

(a)

A random sample of n = 25 specimens is selected.

Compute the probability that the sample average sediment density is at most 3.00 as follows:

Apply continuity correction:

$P(\bar X\leq 3.00)=P(\bar X$

$=P(\bar X$

$=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

$=P(Z$

Thus, the probability that the sample average sediment density is at most 3.00 is 0.092.

Compute the probability that the sample average sediment density is between 2.70 and 3.00 as follows:

$P(2.70$

$=P(0$

*Use a z-table.

Thus, the probability that the sample average sediment density is between 2.70 and 3.00 is 0.477.

(b)

It is provided that:

$P(\bar X\leq 3.00)\geq 0.99$

$P(\bar X$

$P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}$

$P(Z$

The value of z for the probability above is

z ≥ 2.33

Compute the value of n as follows:

$\frac{|2.50-2.70|}{0.75/\sqrt{n}}\geq 2.33$

$\frac{|-0.20|}{2.33}\geq \frac{0.75}{\sqrt{n}}$

$\sqrt{n}\geq \frac{0.75\times 2.33}{|-0.20|}\\\sqrt{n}\geq 8.7375\\n\geq 76.3439\\\approx n\geq 77$

Thus, the sample size must be at least 77.

8 0

3 years ago