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3 years ago

10

Please help with #22! You don’t have to fully explain.

1 answer:

trasher [3.6K]3 years ago

3 0

The answer to the first part is 3 1/2 and the answer to the second question is 2/7

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Describe and correct the error in telling whether the two figures are congruent?

Aleks04 [339]

2 figures are congruent if at least 2 sides are the same length, there angles are the same, etc

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2 years ago

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2 thirds of a number plus 5 is greater than 12

Amanda [17]

Answer:

2/3 of n +5>12

Step-by-step explanation:

6 0

2 years ago

Solve the following:<br> a) 4x-3 = 2x + 7<br><br> b)<br> 2x+6=7x14

Igoryamba

$\huge \qquad \qquad \sf \underline{Answer}$

$\textbf {Let's Solve :}$

<h3>Question 1 :</h3>

$\textsf{4x - 3 = 2x +7}$

$\textsf{4x -2x = 3 +7}$

$\textsf{2x = 10}$

$\textsf{x=5}$

<h3>Question 2 :</h3>

$\sf{2x + 6 = 7 \times 14}$

$\sf{2x + 6 = 98}$

$\sf{2x = 98-6}$

$\sf{2x = 92}$

$\sf{x=46}$

6 0

1 year ago

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

Elena L [17]

Answer:

The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

∴ a = 5 unit

Similarly

The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

Now, area of Triangle written as , from Heron's formula

A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

I.e s = $\frac{5+2+\sqrt{13}}{2}$

Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$

Or, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}$

Or, A = $\frac{3}{2}$ × $\sqrt{1+\sqrt{13} }$

∴ Area of triangle = 1.5 $\sqrt{4.6}$

Hence The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$ Answer

7 0

3 years ago

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Solve the following inequality: 9x+2/3(−7x+6)−1≥2

klasskru [66]

Answer:

Step-by-step explanation:

3(9x - 14/3x + 4 - 1 ≥ 2)

27x - 14x + 12 - 3 ≥ 6

13x + 9 ≥ 6

13x ≥ -3

x ≥ -3/13

5 0

3 years ago

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