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3 years ago

7

What is 3r plus 3s in ditributed property

1 answer:

Nataliya [291]3 years ago

6 0

3r + 3s = 3(r + s) answer

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Find the 67th term of the arithmetic sequence –17, -23, -29,...

Maksim231197 [3]

Answer:

T₆₇ = -413

Step-by-step explanation:

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3 years ago

Read 2 more answers

Bonita deposited 1300 into a bank account that earned 5.75% simple interest each year. She earned $299 in interest before closin

worty [1.4K]

Answer:

For 4 years the money was in the account .

Step-by-step explanation:

Formula

$Simple\ interest = \frac{Principle\times Rate\times Time}{100}$

As given

Bonita deposited 1300 into a bank account that earned 5.75% simple interest each year.

She earned $299 in interest before closing the account.

Principle = $1300

Rate = 5.75%

Simple interest = $299

Putting all the values in the formula

$299 = \frac{1300\times 5.75\times Time}{100}$

$Time= \frac{299\times 100}{1300\times 5.75}$

$Time= \frac{29900\times 100}{1300\times 575}$

$Time= \frac{2990000}{747500}$

Time = 4 years

Therefore for 4 years the money was in the account .

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3 years ago

Math:<br> Please help me. I don’t understand. Thank you for your time.

GuDViN [60]

I Believe it’s 1,890

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3 years ago

Points P and Q belong to segment AB . If AB = a, AP = 2PQ = 2QB, find the distance: midpoints between AP QB

Digiron [165]

Answer: The distace between midpoints of AP and QB is $\frac{a}{8}$ .

Step-by-step explanation: Points P and Q are between points A and B and the segment AB measures a, then:

AP + PQ + QB = a

According to the question, AP = 2 PQ = 2QB, so:

PQ = $\frac{AP}{2}$

QB = $\frac{AP}{2}$

Substituing:

AP + 2*( $\frac{AP}{2}$ ) = a

2AP = a

AP = $\frac{a}{2}$

Since the distance is between midpoints of AP and QB:

2QB = AP

QB = $\frac{AP}{2}$

QB = $\frac{a}{2}*\frac{1}{2}$

QB = $\frac{a}{4}$

MIdpoint is the point that divides the segment in half, so:

<u>Midpoint of AP</u>:

$\frac{AP}{2} = \frac{a}{2}*\frac{1}{2}$

$\frac{AP}{2} = \frac{a}{4}$

<u>Midpoint of QB</u>:

$\frac{QB}{2} = \frac{a}{4}*\frac{1}{2}$

$\frac{QB}{2} = \frac{a}{8}$

The distance is:

d = $\frac{a}{4} - \frac{a}{8}$

d = $\frac{a}{8}$

4 0

3 years ago

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of te triangle is 16m to the power of 2, what are the base

Olenka [21]

Answer:

The base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

Step-by-step explanation:

Given:

A triangle has a base of x 1/2 m and a height of x 3/4 m. If the area of the triangle is 16m to the power of 2.

Now, to find the base and height of the triangle.

The base of triangle = $x\times\frac{1}{2} =\frac{x}{2} \ m.$

The height of triangle = $x\times \frac{3}{4} =\frac{3x}{4}\ m.$

The area of triangle = $16\ m^2.$

Now, we put the formula of area to solve:

$Area=\frac{1}{2} \times base\times height$

$16=\frac{1}{2} \times \frac{x}{2} \times \frac{3x}{4}$

$16=\frac{3x^2}{16}$

<em>Multiplying both sides by 16 we get:</em>

<em /> $256=3x^2$ <em />

<em>Dividing both sides by 3 we get:</em>

<em /> $\frac{256}{3} =x^2$ <em />

<em>Using square root on both sides we get:</em>

$\frac{16}{\sqrt{3}}=x$

$x=\frac{16}{\sqrt{3}}$

Now, by substituting the value of $x$ to get the base and height:

$Base=\frac{x}{2}\\\\Base=\frac{\frac{16}{\sqrt{3}}}{2} \\\\Base=\frac{8}{\sqrt{3}} \ m.$

<em>So, the base of triangle = </em> $\frac{8}{\sqrt{3}} \ m.$ <em />

$Height=x\times\frac{3}{4} \\\\Height=\frac{16}{\sqrt{3}}\times \frac{3}{4} \\\\Height=\frac{12}{\sqrt{3}} \ m.$

<em>Thus, the height of triangle = </em> $\frac{12}{\sqrt{3}} \ m.$ <em />

Therefore, the base of triangle is $\frac{8}{\sqrt{3}} \ m$ and the height of triangle is $\frac{12}{\sqrt{3}} \ m.$

7 0

3 years ago