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2 years ago

13

Is 0.83 equal to 83 over 100

1 answer:

dedylja [7]2 years ago

3 0

Yes because 83 divided by 100 is .83

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The lines below are perpendicular if the slope of the green line is 3/2 what is the slope of the red line

shepuryov [24]

Answer:

The perpendicular line has a slope of -2/3

Step-by-step explanation:

Perpendicular lines have a slope that multiplies to -1

3/2 * m = -1

Multiply each side by 2/3

3/2 * 2/3 *m = -1 * 2/3

m = -2/3

8 0

2 years ago

Read 2 more answers

Pls help I'm bad at math v.v<br><br> Lines m and n are parallel.

Eva8 [605]

Answer:

The measure of angle 1 is 55.

Step-by-step explanation:

The top angle= 75 because of the vertical angle theorem

The right angle= 50 because of the alternate exterior angle theorem

75+50=125

There are 180 degrees in a triangle so the angle has to equal 55

5 0

3 years ago

The area of a parallelogram is 6 square units. Both the height and the length of the base are whole numbers. What are the possib

lbvjy [14]

Answer:

Step-by-step explanation:

The factor pairs of 6 are 1×6 and 2×3

So the parallelogram (height by length) could be:

1 by 6

2 by 3

3 by 2

6 by 1

7 0

2 years ago

What is the area of this parallelogram?

Lorico [155]

Step-by-step explanation:

<u>Parallelogram formula:</u> $A=b(base)h(height)$

<u>The value </u><u>6</u><u> is meant to trick </u><u>you</u><u>.</u>

The height is 7, and the base is 10 1/4.

<u>Multiply the base and height:</u>

<u /> $7 \times 10\frac{1}{4} = 71\frac{3}{4}$

The area equals: $71\frac{3}{4}$ $ft^{2}$

4 0

1 year 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

2 years ago