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3 years ago

15

A painter can finish painting a house in 7 hours. Her assistant takes 9 hours to finish the same job. How long would it take for

them to complete the job if they were working together?

2 answers:

Vesnalui [34]3 years ago

6 0

Answer:

3.9375 hours

Step-by-step explanation:

Combined time

= product of individual times ÷ sum

= (7×9)/(7+9)

= 63/16

= 3.9375

andre [41]3 years ago

5 0

Answer:

c =3.9375 hours

Step-by-step explanation:

The formula to determine time to finish the job together is

1/a + 1/b = 1/c

Where a and b are the times to complete the job separately and c is the time to complete the job when working together

1/7 + 1/9 = 1/c

Multiply by 63c to get rid of the fractions

63c*(1/7 + 1/9) = 1/c *63c

9c + 7c = 63

Combine like terms

16c = 63

Divide each side by 16

16c/16 = 63/16

c =3.9375 hours

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3 years ago

Solve for all values of x by factoring. x^2+3x+8=-3x+3

docker41 [41]

Answer:

x = -1,-5

Step-by-step explanation:

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3 years ago

Solve the equation 5x+2-x=-4x

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8 0

3 years ago

Read 2 more answers

Janice wants to buy carpet for a trapezoid shaped room. The bases of the trapezoid are 12 feet and 14 feet, and the height is 15

Usimov [2.4K]

For this case, the first thing to do is find the area of the trapezoid.
We have then:
$A = (1/2) * (b1 + b2) * (h)
$
Where,
b1: b2: Trapezoid bases
h: height of the trapezoid
Substituting values we have:
$A = (1/2) * (12 + 14) * (15)

A = 195 ft ^ 2$
Then, the cost per square foot is given by:
$C = 5.50A

C = 5.50 * 195

C = 1072.5$
Answer:
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C = 1072.5 $

4 0

3 years ago

The expression (secx + tanx)2 is the same as _____.

trapecia [35]

<u>Answer:</u>

The expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

<u>Solution:</u>

From question, given that $\bold{(\sec x+\tan x)^{2}}$

By using the trigonometric identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$ the above equation becomes,

$(\sec x+\tan x)^{2} = \sec ^{2} x+2 \sec x \tan x+\tan ^{2} x$

We know that $\sec x=\frac{1}{\cos x} ; \tan x=\frac{\sin x}{\cos x}$

$(\sec x+\tan x)^{2}=\frac{1}{\cos ^{2} x}+2 \frac{1}{\cos x} \frac{\sin x}{\cos x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

$=\frac{1}{\cos ^{2} x}+\frac{2 \sin x}{\cos ^{2} x}+\frac{\sin ^{2} x}{\cos ^{2} x}$

On simplication we get

$=\frac{1+2 \sin x+\sin ^{2} x}{\cos ^{2} x}$

By using the trigonometric identity $\cos ^{2} x=1-\sin ^{2} x$ ,the above equation becomes

$=\frac{1+2 \sin x+\sin ^{2} x}{1-\sin ^{2} x}$

By using the trigonometric identity $(a+b)^{2}=a^{2}+2ab+b^{2}$

we get $1+2 \sin x+\sin ^{2} x=(1+\sin x)^{2}$

$=\frac{(1+\sin x)^{2}}{1-\sin ^{2} x}$

$=\frac{(1+\sin x)(1+\sin x)}{1-\sin ^{2} x}$

By using the trigonometric identity $a^{2}-b^{2}=(a+b)(a-b)$ we get $1-\sin ^{2} x=(1+\sin x)(1-\sin x)$

$=\frac{(1+\sin x)(1+\sin x)}{(1+\sin x)(1-\sin x)}$

$= \frac{1+\sin x}{1-\sin x}$

Hence the expression $\bold{(\sec x+\tan x)^{2} \text { is same as } \frac{1+\sin x}{1-\sin x}}$

8 0

3 years ago