Ask question

Ask question

All categories

katovenus [111]

2 years ago

11

10) what is the area ratio 11) what is the area of RST HELP ASAP EMERGENCY

1 answer:

dlinn [17]2 years ago

8 0

9514 1404 393

Answer:

10) C 4/9

11) B 24

Step-by-step explanation:

10) The area ratio is the square of the ratio of side lengths. The ratio of side lengths is ...

RS/LM = 6/9 = 2/3

Then the area ratio is ...

(2/3)² = 4/9

__

11) Using the result of question 10, the area of ∆RST is ...

(4/9)(54 in²) = 24 in²

You might be interested in

7/10 + 2/10 , 13/16+ 2/16 , 4/5 + 1/5 , 7/15 + 2/15 , 9/20 + 3/20 , 5/8 + 1/8 . All in simplist form.

blondinia [14]

Hello!

<h3><em><u>Answers</u></em></h3>

1. $\frac{7}{10} + \frac{2}{10} = \frac{9}{10}$ Simplest form: $\frac{9}{10}$

2. $\frac{13}{16} + \frac{2}{16} = \frac{15}{16}$ Simplest form: $\frac{15}{16}$

3. $\frac{4}{5} + \frac{1}{5} = \frac{5}{5}$ Simplest form: $\frac{1}{1}$

4. $\frac{7}{15} + \frac{2}{15} = \frac{9}{15}$ Simplest form: $\frac{3}{5}$

5. $\frac{9}{20} + \frac{3}{20} = \frac{12}{20}$ Simplest form: $\frac{3}{5}$

6. $\frac{5}{8} + \frac{1}{8} = \frac{6}{8}$ Simplest form: $\frac{3}{4}$

<h3><em><u>Explanation:</u></em></h3>

Simply add the numerators of the like fractions together. The denominators remain the same.

4 0

3 years ago

If i= the square root of -1, what is the value of i to the 3rd power

serious [3.7K]

I^3 = i^(2+1) = i^2 x i -1 x i = -i

3 0

3 years ago

What is the period of the sinusoidal function?

GuDViN [60]

Answer:

i dont know my guy

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Hello I need help with this question

Andreyy89

The answer is x=7. Since the two angles are corresponding. Then it would be 10x-2=9x+5. The answer would be x=7. Please mark me Brainleist

7 0

2 years ago

A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set

lara [203]

A circle with center $(3,0)$ and radius $5$ has equation

$(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x = 16$

If we substitute $y=2x+k$ in this equation, we have

$x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0$

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

$\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0$

The expression simplifies to

$-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89$

The solutions to the associated equation are

$k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}$

So, the parabola is negative between the two solutions:

$-6-5\sqrt{5}$

8 0

3 years ago