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3 years ago

9

Can I get some help with question 12? I need to solve it while making a proof. I’ve asked at least 5 tutors for help and none of

them know what to do.

1 answer:

Oliga [24]3 years ago

3 0

Answer:

Step-by-step explanation:

Angle ABD is congruent to angle BDC by alternate interior angles of the parallel sides AB and DC.

Angle GEB is a right angle by given.

Angle DFC is a right angle by given.

The remaining angles for each triangle are congruent by sum of angles of a triangle is 180 degrees.

The two triangles are similar by angle-angle-angle.

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What are 2 square numbers that have a sum of 130

kenny6666 [7]

The square roots are 7 and 9 so the square numbers would be 49 and 81 to equal 130

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3 years ago

Which are solutions of the equation (x + 5)(x – 3) = 0? Check all that apply

meriva

-3+5x(-3)-3=0. So -3 is the answer

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3 years ago

Solve the inequality, H-16<_-24

Semenov [28]

H - 16 <span>≤ -24
h </span>≤ -24 + 16
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5 0

4 years ago

Read 2 more answers

-8(6+5x) = 3n - 5 <br><br> Help please

BaLLatris [955]

$\large\green{\tt Correct \: Question :-}$

$\to\tt -8(6+5x) = 3x - 5$

$\large\gray{\tt Answer :-}$

$\implies$ - 1

$\large\blue{\tt Given :-}$

$\to\tt -8(6+5x) = 3x - 5$

$\large\orange{\tt To \: Find :-}$

$\to\tt Value \: of \: \boxed{\bf x}$

$\large\red{\tt Solution :-}$

$:\implies\tt -8(6+5x) = 3x - 5$

$:\implies\tt - 48 - 40x = 3x - 5$

$:\implies\tt - 40x - 3x = - 5 + 48$

$:\implies\tt - 43x = 43$

$:\implies\tt x = \frac{\cancel{43}}{\cancel{-43}}$

$:\implies\tt x = \frac{\cancel1}{\cancel{- 1}}$

$:\implies\tt x = -1$

$\large\pink{\tt Verification :-}$

We have,

$:\implies\tt -8(6+5x) = 3x - 5$

Substituting the value of x,

$:\implies\tt - 48 - 40 \times ( - 1) = 3 \times ( - 1) - 5$

$:\implies\tt - 48 + 40 = - 3 - 5$

$:\implies\tt - 8 = - 8$

<h3><u>HENCE VERIFIED</u></h3>

4 0

3 years ago

Help please! these questions are really confusing me

Anettt [7]

Answer:

Figure 1: Option A : $\sqrt{\frac{\textbf{3}}{\textbf{2}}}$ ; $\frac{\textbf{1}}{\textbf{2}}$ ; $\sqrt{\textbf{3}}}$

Figure 2: Option B: $\frac{\sqrt{\textbf{19}}}{\textbf{10}}$ ; $\frac{\textbf{9}}{\textbf{10}}$ ; $\frac{\sqrt{\textbf{19}}}{\textbf{9}}$

Figure 3: Option C: $\frac{\textbf{4}}{\textbf{5}}$ ; $\frac{\textbf{3}}{\textbf{5}}$ ; $\frac{\textbf{4}}{\textbf{3}}$

Step-by-step explanation:

$\textbf{Sin A} \hspace{1mm} \textbf{= } \hspace{1mm} \frac{\textbf{opp}}{\textbf{hyp}}$

$\textbf{Cos A} \hspace{1mm} {\textbf{=} \hspace{1mm} \frac{\textbf{adj}}{\textbf{hyp}}$

$\textbf{Tan A} \hspace{1mm} \textbf{=} \hspace{1mm} \frac{\textbf{Sin A}}{\textbf{Cos A}}$

We can simply follow these three formulas to solve the problem.

Figure 1:

Sin A = $\frac{6\sqrt{3}}{12}$ = $\frac{\sqrt{3}}{2}$

Cos A = $\frac{6}{12} = \frac{1}{2}$

Tan A = $\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}}$ = $\sqrt{3}$

Figure 2:

Sin A = $\frac{2\sqrt{19}}{20}$ = $\frac{\sqrt{19}}{10}$

Cos A = $\frac{18}{20} = \frac{9}{10}$

Tan A = $\frac{\frac{\sqrt{19}}{10}}{\frac{10}{9}}$ = $\frac{\sqrt{19}}{9}$

Figure 3:

Sin A = $\frac{16}{20}$ = $\frac{4}{5}$

Cos A = $\frac{12}{20} = \frac{3}{5}$

Tan A = $\frac{\frac{4}{5}}{\frac{3}{5}}$ = $\frac{4}{3}$

Hence, the answer.

6 0

3 years ago