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3 years ago

One of two or more Expressions that are multiplied together to get a product

Mathematics

1 answer:

slamgirl [31]3 years ago

6 0

Answer:

factor

Step-by-step explanation:

We assume this is a vocabulary question and you want the word that has the definition you have supplied.

Such an expression is called a <em>factor</em>.

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1. Carter will install fencing all around the flat area of his

MariettaO [177]

Answer:

Amount of fencing required = 75 yards

Step-by-step explanation:

Distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by the formula,

d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between A(3, 6) and B(3, -2) = $\sqrt{(3-3)^2+(6+2)^2}$

= 8 yards

Distance between B(3, -2) and C(-7, 4) = $\sqrt{(3+7)^2+(-2-4)^2}$

= $\sqrt{136}$

= 11.66 ≈ 12 yards

Distance between C(-7, 4) and D(-7, -2) = $\sqrt{(-7+7)^2+(4+2)^2}$

= 6 yards

Distance between D(-7, -2) and E(-3, -2) = $\sqrt{(-7+3)^2+(-2+2)^2}$

= 4 yards

Distance between E(-3, -2) and F(-3, -8) = $\sqrt{(-3+3)^2+(-2+8)^2}$

= 6 yards

Distance between F(-3, -8) and G(3, -8) = $\sqrt{(-3-3)^2+(-8+8)^2}$

= 6 yards

Distance between G(3, -8) and H(10, -12) = $\sqrt{(3-10)^2+(-8+12)^2}$

= $\sqrt{49+16}$

= $\sqrt{65}$

= 8.06 ≈ 8 yards

Distance between H(10, -12) and J(10, 6) = $\sqrt{(10-10)^2+(-12-6)^2}$

= 18 yards

Distance between A(3, 6) and J(10, 6) = $\sqrt{(10-3)^2+(6-6)^2}$

= 7 yards

Since length of fence required = perimeter of the flat area

Perimeter of the given area = 8 + 12 + 6 + 4 + 6 + 6 + 8 + 18 + 7

= 75 yards

Therefore, amount of fencing required = 75 yards

8 0

3 years ago

What property was applied to solve the equation below? Negative StartFraction 3 over 8 EndFraction x = 48 Negative StartFraction

Soloha48 [4]

Multiplication property of equality

Step-by-step explanation:

We are given that the equation

$-\frac{3}{8}x=48$

We have to find the property which was applied to solve the equation.

$-\frac{3}{8}x=48$

Reason: Given

Multiply by -8/3 on both sides

$-\frac{8}{3}\times -\frac{3}{8}x=-\frac{8}{3}\times 48$

By using multiplication property of equality.

$x=\frac{-384}{3}$

After simplification

$x=-128$

4 0

3 years ago

Read 2 more answers

Justin built a display stand out of two rectangular prisms. The smaller rectangular prism is 2 inches tall, 4 inches wide, and 6

Lady bird [3.3K]

Answer:

428 square inches

Step-by-step explanation:

Hi, to answer this question we have to apply the next formula:

Surface area of a prism: 2wl +2lh+2hw

Where w is the width of the prism; l the length; and h the height.

Replacing with the values given:

S (smaller prism) = 2 (4x6) + 2(6x2)+ 2(2x4)=48+24+16= 88 in2

S (larger prism) = 2 (8x10) +2 (10 x5)+2(5x8)= 160+100+80=340in2

Adding both surface areas:

88+340 =428 in2

Feel free to ask for more if needed or if you did not understand something.

5 0

4 years ago

Jeremy had 3/4 of a submarine sandwich and gave

NeX [460]

Actual question is "<span>Jeremy had 3/4 of a submarine sandwich and gave his friend 1/3 of it. What fraction of the sandwich did the friend receive?"

Solution:
1/3 of 3/4 of a submarine sandwich = 1/3x3/4 = 3/12 = 1/4
so jermy gave 1/4 of the submarine sandwich to his friend. </span>

5 0

3 years ago

A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the l

Anna [14]

Answer:

a) 0.4452

b) 0.0548

c) 0.0501

d) 0.9145

e) 6.08 minutes or greater

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 4.7 minutes

Standard Deviation, σ = 0.50 minutes.

We are given that the distribution of length of the calls is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(calls last between 4.7 and 5.5 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{4.7 - 4.7}{0.50} \leq z \leq \displaystyle\frac{5.5-4.7}{0.50}) = P(0 \leq z \leq 1.6)\\\\= P(z \leq 1.6) - P(z$

$P(4.7 \leq x \leq 5.5) = 44.52\%$

b) P(calls last more than 5.5 minutes)

$P(x > 5.5) = P(z > \displaystyle\frac{5.5-4.7}{0.50}) = P(z > 1.6)\\\\P( z > 1.6) = 1 - P(z \leq 1.6)$

Calculating the value from the standard normal table we have,

$1 - 0.9452 = 0.0548 = 5.48\%\\P( x > 5.5) = 5.48\%$

c) P( calls last between 5.5 and 6 minutes)

$P(4.7 \leq x \leq 5.5) = P(\displaystyle\frac{5.5 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(1.6 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$

$P(5.5 \leq x \leq 6) = 5.01\%$

d) P( calls last between 4 and 6 minutes)

$P(4 \leq x \leq 6) = P(\displaystyle\frac{4 - 4.7}{0.50} \leq z \leq \displaystyle\frac{6-4.7}{0.50}) = P(-1.4 \leq z \leq 2.6)\\\\= P(z \leq 2.6) - P(z$