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steposvetlana [31]

2 years ago

12

Some please help and explain the steps. Tysmmmmm

1 answer:

-Dominant- [34]2 years ago

5 0

Answer:

since angle B and angle C contains the same arc AD. set the angles equal

11x-3=8x+15

x=6

so the angle has a measure of 63 degrees

the central angle is the angle of the arc

and it is always 2 times the angle that we just that contains the arc but does touches the end of the circle.

so its 63 *2 =126 degrees.

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Given the system below, which points are solutions? (check all that apply)

LuckyWell [14K]

It’s 3.46 and 6.64hrhss

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3 years ago

A radius is an angle that connects any point on the circle to the center of that circle.

Flura [38]

The answer is false.

A radius is a segment that connects any point on the circle to the center of said circle. An angle requires two lines, and a radius only consists of one line, which further proves that this statement is false.

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3 years ago

22. Find the perimeter and area.

zmey [24]

Answer:

Part 22) The area is $A=15a^3b^6\ units^2($ and the perimeter is $P=10a^2b^4+6ab^2\ unit$

Part 24) The area is $A=16m^3n\ units^2$ and the perimeter is $P=24mn\ units$

Part 26) The area is equal to $A=9\pi x^6y^{2}\ units^2$

Step-by-step explanation:

Part 22) Find the perimeter and area

step 1

The area of a rectangle is equal to

$A=LW$

we have

$L=5(a^2)(b^4)\ units$

$W=3(a)(b^2)\ units$

Remember that

When multiply exponents with the same base, adds the exponents and maintain the base

substitute in the formula

$A=(5(a^2)(b^4))(3(a)(b^2))$

$A=15a^3b^6\ units^2$

step 2

The perimeter of a rectangle is equal to

$P=2(L+W)$

we have

$L=5(a^2)(b^4)\ units$

$W=3(a)(b^2)\ units$

substitute in the formula

$P=2(5(a^2)(b^4)+3(a)(b^2))$

$P=10a^2b^4+6ab^2\ unit$

Part 24) Find the perimeter and area

step 1

The area of triangle is equal to

$A=\frac{1}{2}bh$

where

$b=8mn\ units$

$h=4m^2\ units$

Remember that

When multiply exponents with the same base, adds the exponents and maintain the base

substitute the given values

$A=\frac{1}{2}(8mn)(4m^2)$

$A=16m^3n\ units^2$

step 2

Find the perimeter

I will assume that is an equilateral triangle (has three equal length sides)

The perimeter of an equilateral triangle is

$P=3b$

where

$b=8mn\ units$

substitute

$P=3(8mn)$

$P=24mn\ units$

Part 26) Find the area

The area of a circle is equal to

$A=\pi r^{2}$

where

$r=3x^3y\ units$

Remember the property

$(a^{m})^{n}=a^{m*n}$

substitute in the formula the given value

$A=\pi (3x^3y)^{2}$

$A=9\pi x^6y^{2}\ units^2$

6 0

3 years ago

What is y equals negative three fourths plus two equal

Crazy boy [7]

Y = -3/4 + 2

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7 0

3 years ago

Drag the tiles to the boxes to form correct pairs.<br> Match the pairs of equivalent expressions.

Rashid [163]

Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

<em>Solving the expression</em>

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

<em>Solving the expression</em>

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

<em>Solving the expression</em>

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

As

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

<em>Solving the expression</em>

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

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5 0

3 years ago

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