When adding the square root of 9 and -7, which type of number is the sum?

What is the value of the expression? 9^2÷(-3)^0

stiv31 [10]

Step 1. Simplify 9^2 to 81
81 ÷ (-3)^0
Step 2. Use Rule of Zero: x^0 = 1
81 ÷ 1
Step 3. Simplify
81

5 0

3 years ago

the area of a rectangle is 90 in2^. the ratio of the length to the width is 5:2. find the length and the width

-BARSIC- [3]

Length and width of rectangle is 15 inches and 6 inches respectively

<h3>Solution:</h3>

Given that area of a rectangle is 90 square inch

Ratio of length to the width = 5: 2.

Need to determine length and width of rectangle.

As ratio of length to the width is 5 : 2

Lets assume length of rectangle = 5x inches and width of rectangle = 2x inches.

The formula for area of rectangle is given as:

$\text { Area of rectangle }=\text { length of rectangle } \times \text { width of rectangle}$

Substituting the given value of area of rectangle and assumed value of length and width of rectangle we get:

$\begin{array}{l}{90=5 x \times 2 x} \\\\ {=>90=10 x^{2}}\end{array}$

On solving the above expression for x we get

$\begin{array}{l}{=>\frac{90}{10}=x^{2}} \\\\ {=>x^{2}=9} \\\\ {=>x=\sqrt{9}=3}\end{array}$

$\begin{array}{l}{\text { Length of rectangle }=5 \times x=5 \times 3=15 \text { inches }} \\\\ {\text { Width of rectangle }=2 \times} x=2 \times} 3=6 \text { inches }}\end{array}$

Hence length and width of rectangle is 15 inches and 6 inches.

4 0

4 years ago

What’s the triangle congruence shortcut

mina [271]

Answer:

D. All those resources can be found in Africa

7 0

3 years ago

Read 2 more answers

How do you find the formula of a half circle?

Alik [6]

Full circle
Circumference = 2*pi * r
Circumference = pi * d

Area = pi r * r = pi * r^2
Area = pi d^2 / 4

Half circle
Circumference = pi * r
Circumference = pi * r / 2

Area = pi r^2 / 2
Area = pi d^2/8

The answer depends on what you are looking for and what you are given. But those are the only common formulas I can think of at the moment.

4 0

3 years ago

Read 2 more answers

Evaluate cos(sin^-1(4/5)

PSYCHO15rus [73]

Step-by-step explanation:

$Let \: \: \theta = \sin^{ - 1} (\frac{4}{5}) \\ \\ \therefore \: \sin\theta = \frac{4}{5} \\ \\ \because \: { \cos}^{2} \theta =1 - { \sin}^{2} \theta \\ \\ \therefore \: { \cos}^{2} \theta = 1 - (\frac{4}{5} )^{2} \\ \\ = 1 - \frac{16}{25} \\ \\ = \frac{25 - 16}{25} \\ \\ = \frac{9}{25} \\ \\ \therefore \:{ \cos}\theta = \pm \: \frac{3}{5} \\ \\ \because \: \theta \: lie \: in \: the \: first \: quadrant \\ \\ \therefore \: { \cos}\theta = \: \frac{3}{5} \\ \\ \implies \theta = { \cos}^{ -1 } \: \frac{3}{5}\\\\\implies \theta = { \cos}^{ -1 } \: \frac{3}{5}= { \sin}^{ -1 } \: \frac{4}{5} \\ \\ \therefore \: \cos( \ {sin}^{ - 1} \frac{4}{5} ) \\ \\ = \cos( \ {cos}^{ - 1} \frac{3}{5} ) \\\\ = \frac{3}{5} \\ \\ \purple{ \boxed{\therefore \: \cos( \ {sin}^{ - 1} \frac{4}{5} ) = \frac{3}{5}}}$

8 0

3 years ago