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Amiraneli [1.4K]

3 years ago

12

Convert 0.0000001 to a power of 10. 107 10−7 106 10−6

2 answers:

valina [46]3 years ago

4 0

Since you have to move the decimal point the the right by 7 places to get the decimal point after the one. You must multiply by 1/10 seven times, which is the same as:

10^-7

defon3 years ago

4 0

10^-7 = 0.0000001

when the number is a decimal the power is a negative number

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Plz help me I will mark u Brainiest

Reika [66]

Answer: 1- quadrant 1 or top right

2- quadrant 4 or bottom right

3-quadrant 3 or bottom left

4 0

3 years ago

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∠A and ∠B are supplementary, and ∠A and ∠C are supplementary. Which conclusion is valid? Select one: A. ∠B and ∠C are supplement

zysi [14]

Option D is the correct answer.

Answer:

D. ∠B and ∠C are congruent.

Step-by-step explanation:

Since, ∠A and ∠B are supplementary.

Therefore,

∠A + ∠B = 180°.....(1)

Since, ∠A and ∠C are supplementary.

Therefore,

∠A + ∠C = 180°.....(2)

From equations (1) & (2)

∠A + ∠B = ∠A + ∠C

=> ∠B = ∠C

Hence, ∠B and ∠C are congruent.

4 0

4 years ago

Rewrite y + 1 = -2x – 3 in standard form

Misha Larkins [42]

The algebraic expressions can be written as

$a+b+c=0$

The given expression is,

$\begin{gathered} y+1=-2x-3 \\ -2x-y=1+3 \\ -2x-y=4 \\ -2x-y-4=0 \\ 2x+y+4=0 \end{gathered}$

7 0

1 year ago

What is the degree of the following polynomial? 4x^2+2x+7

adelina 88 [10]

The degree of the polynomial is 2.

to find the degree, put the polynomial in descending order with the highest exponents in front. the highest exponent is the degree.

3x^6 - 4 (highest exponent here is 6, degree would be 6)

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3 years ago

The vertices of ∆ABC are A(-2, 2), B(6, 2), and C(0, 8). The perimeter of ∆ABC is units is? What is the area?

11111nata11111 [884]

we have

$A(-2, 2),B(6, 2),C(0, 8)$

see the attached figure to better understand the problem

we know that

The perimeter of the triangle is equal to

$P=AB+BC+AC$

and

the area of the triangle is equal to

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB\\heigth=DC$

we know that

The distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step $1$

<u>Find the distance AB</u>

$A(-2, 2),B(6, 2)$

Substitute the values in the formula

$d=\sqrt{(2-2)^{2}+(6+2)^{2}}$

$d=\sqrt{(0)^{2}+(8)^{2}}$

$dAB=8\ units$

Step $2$

<u>Find the distance BC</u>

$B(6, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-6)^{2}}$

$d=\sqrt{(6)^{2}+(-6)^{2}}$

$dBC=6\sqrt{2}\ units$

Step $3$

<u>Find the distance AC</u>

$A(-2, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0+2)^{2}}$

$d=\sqrt{(6)^{2}+(2)^{2}}$

$dAC=2\sqrt{10}\ units$

Step $4$

<u>Find the distance DC</u>

$D(0, 2),C(0, 8)$

Substitute the values in the formula

$d=\sqrt{(8-2)^{2}+(0-0)^{2}}$

$d=\sqrt{(6)^{2}+(0)^{2}}$

$dDC=6\ units$

Step $5$

<u>Find the perimeter of the triangle</u>

$P=AB+BC+AC$

substitute the values

$P=8\ units+6\sqrt{2}\ units+2\sqrt{10}\ units$

$P=22.81\ units$

therefore

The perimeter of the triangle is equal to $22.81\ units$

Step $6$

<u>Find the area of the triangle</u>

$A=\frac{1}{2}*base *heigth$

in this problem

$base=AB=8\ units\\heigth=DC=6\ units$

substitute the values

$A=\frac{1}{2}*8*6$

$A=24\ units^{2}$

therefore

the area of the triangle is $24\ units^{2}$

4 0

3 years ago

Read 2 more answers