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erastova [34]
2 years ago
9

You work 4 hours on Monday, 4.5 hours on Tuesday, 7.25 hours on Thursday, and 12 hours on Saturday. If you get paid $11.52 per h

our, what is your total pay for the week?
Mathematics
1 answer:
likoan [24]2 years ago
7 0

Answer:

319.68

Step-by-step explanation:

you multyply 11.52 per each hour and then add everything up

Wish this helped you

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Draw a line representing the "rise" and a line representing the "run" of the line. State
AleksAgata [21]
The answer would be -3/5. The slope is the rise over run. It goes up 3 and over 5. Hope this helps!
3 0
2 years ago
Read 2 more answers
How many radians is 210
Rina8888 [55]

Answer:

7pi/6 radians

Step-by-step explanation:

To change to radians multiply by pi/180

210 * pi/180

210 pi/180

7pi/6

5 0
2 years ago
Read 2 more answers
Please show the working and answer. you can take a picture for the working.
baherus [9]

Answer:

(a) The area of the triangle is approximately 39.0223 cm²

(b) ∠SQR is approximately 55.582°

Step-by-step explanation:

(a) By sin rule, we have;

SQ/(sin(∠SPQ)) = PQ/(sin(∠PSQ)), which gives;

5.4/(sin(52°)) = 6.8/(sin(∠PSQ))

∴ (sin(∠PSQ)) = (6.8/5.4) × (sin(52°)) ≈ 0.9923

∠PSQ = sin⁻¹(0.9923) ≈ 82.88976°

Similarly, we have;

5.4/(sin(52°)) = SP/(sin(180 - 52 - 82.88976))

Where, 180 - 52 - 82.88976 = ∠PQS = 45.11024

SP = 5.4/(sin(52°))×(sin(180 - 52 - 82.88976)) ≈ 4.8549

Given that RS : SP = 2 : 1, we have;

RS = 2 × SP = 2 × 4.8549 ≈ 9.7098

We have by cosine rule, \overline {RQ}² =  \overline {SQ}² +  \overline {SR}² - 2 × \overline {SQ} × \overline {SR} × cos(∠QSR)

∠QSR and ∠PSQ are supplementary angles, therefore;

∠QSR = 180° - ∠PSQ = 180° - 82.88976° = 97.11024°

∠QSR = 97.11024°

Therefore;

\overline {RQ}² =  5.4² +  9.7098² - 2 ×  5.4×9.7098× cos(97.11024)

\overline {RQ}² ≈ 136.42

\overline {RQ} = √(136.42) ≈ 11.6799

The area of the triangle = 1/2 ×\overline {PQ} × \overline {PR} × sin(∠SPQ)

By substituting the values, we have;

1/2 ×\overline {PQ} × \overline {PR} × sin(∠SPQ)

1/2 × 6.8 × (4.8549 + 9.7098) × sin(52°) ≈ 39.0223 cm²

The area of the triangle ≈ 39.0223 cm²

(b) By sin rule, we have;

\overline {RS}/(sin(∠SQR)) = \overline {RQ}/(sin(∠QSR))

By substituting, we have;

9.7098/(sin(∠SQR)) = 11.6799/(sin(97.11024))

sin(∠SQR) = 9.7098/(11.6799/(sin(97.11024))) ≈ 0.82493

∠SQR = sin⁻¹(0.82493) ≈ 55.582°.

8 0
2 years ago
For f(x)=2x+1 and g(x)=x^2-7, find (f+g)(x) A. 2x^2-15 B.X^2+2x-6 C.2x^3-6 D.x^2+2x+8
Advocard [28]

Answer:

C

Step-by-step explanation:

(f + g)(x) = f(x) + g(x) , thus

f(x) + g(x)

= 2x + 1 + x² - 7 ← collect like terms

= x² + 2x - 6 → C

5 0
3 years ago
If the initial amount of iodine-131 is 537 grams , how much is left after 10 days?
viktelen [127]

Answer:

225.78 grams

Step-by-step explanation:

To solve this question, we would be using the formula

P(t) = Po × 2^t/n

Where P(t) = Remaining amount after r hours

Po = Initial amount

t = Time

In the question,

Where P(t) = Remaining amount after r hours = unknown

Po = Initial amount = 537

t = Time = 10 days

P(t) = 537 × 2^(10/)

P(t) = 225.78 grams

Therefore, the amount of iodine-131 left after 10 days = 225.78 grams

4 0
3 years ago
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