All categories

3 years ago

Estimate the quotient using compatible numbers. 2,252 ÷ 7

Mathematics

2 answers:

Fittoniya [83]3 years ago

7 0

321.7. It's rounded down to the appropriate number of figures.

Ede4ka [16]3 years ago

4 0

321.7 is the answer

You might be interested in

Ava buys a book for $8.58. she pays with a $10 bill. how much change will she get?

mestny [16]

It's $1.42 have a nice day

8 0

3 years ago

Read 2 more answers

Help I can’t figure out how to do zero exponents

Orlov [11]

Anything (except 0) to the 0 power is equal to 1.

8^0 = 1

(-8)^0 = 1

24781297429879741^0 = 1

4 0

2 years ago

Solve for x and select your answer from the multiple choices.

Alchen [17]

Answer:

x=3

Step-by-step explanation:

3x+5=14

-5 -5

3x=9

/3 /3

x=3

3 0

3 years ago

Read 2 more answers

The owner of a used car lot wants to buy as many cars at an auction as he can while not spending more than $20,000. if he expect

IRISSAK [1]

Answer:

5 cars

Step-by-step explanation:

If the average of cost of a car is $4,000 there will be enough cash to buy 5 cars. 20,000 divided by 4,000= 5.

7 0

3 years ago

Cosθ=−2√3 , where π≤θ≤3π2 .

Alex787 [66]

Answer:

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

Step-by-step explanation:

step 1

Find the $sin(\theta)$

we know that

Applying the trigonometric identity

$sin^2(\theta)+ cos^2(\theta)=1$

we have

$cos(\theta)=-\frac{\sqrt{2}}{3}$

substitute

$sin^2(\theta)+ (-\frac{\sqrt{2}}{3})^2=1$

$sin^2(\theta)+ \frac{2}{9}=1$

$sin^2(\theta)=1- \frac{2}{9}$

$sin^2(\theta)= \frac{7}{9}$

$sin(\theta)=\pm\frac{\sqrt{7}}{3}$

Remember that

π≤θ≤3π/2

Angle θ belong to the III Quadrant

That means ----> The sin(θ) is negative

$sin(\theta)=-\frac{\sqrt{7}}{3}$

step 2

Find the sec(β)

Applying the trigonometric identity

$tan^2(\beta)+1= sec^2(\beta)$

we have

$tan(\beta)=\frac{4}{3}$

substitute

$(\frac{4}{3})^2+1= sec^2(\beta)$

$\frac{16}{9}+1= sec^2(\beta)$

$sec^2(\beta)=\frac{25}{9}$

$sec(\beta)=\pm\frac{5}{3}$

we know

0≤β≤π/2 ----> II Quadrant

sec(β), sin(β) and cos(β) are positive

$sec(\beta)=\frac{5}{3}$

Remember that

$sec(\beta)=\frac{1}{cos(\beta)}$

therefore

$cos(\beta)=\frac{3}{5}$

step 3

Find the sin(β)

we know that

$tan(\beta)=\frac{sin(\beta)}{cos(\beta)}$

we have

$tan(\beta)=\frac{4}{3}$

$cos(\beta)=\frac{3}{5}$

substitute

$(4/3)=\frac{sin(\beta)}{(3/5)}$

therefore

$sin(\beta)=\frac{4}{5}$

step 4

Find sin(θ+β)

we know that

$sin(A + B) = sin A cos B + cos A sin B$

In this problem

$sin(\theta + \beta) = sin(\theta)cos(\beta)+ cos(\theta)sin (\beta)$

we have

$sin(\theta)=-\frac{\sqrt{7}}{3}$

$cos(\theta)=-\frac{\sqrt{2}}{3}$

$sin(\beta)=\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute the given values in the formula

$sin(\theta + \beta) = (-\frac{\sqrt{7}}{3})(\frac{3}{5})+ (-\frac{\sqrt{2}}{3})(\frac{4}{5})$

$sin(\theta + \beta) = (-3\frac{\sqrt{7}}{15})+ (-4\frac{\sqrt{2}}{15})$

$sin(\theta + \beta) = -\frac{\sqrt{7}}{5}-4\frac{\sqrt{2}}{15}$

8 0

3 years ago