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4 years ago

Is it A,B,C, or D? Please Help. MATH

Mathematics

1 answer:

Solnce55 [7]4 years ago

6 0

In figure B the tetra shape has been rotated 90 degreez

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The number of calories burned while jogging varies directly with the number of minutes spent jogging. If george burns 150 calori

IRISSAK [1]

He will burn 225 calories. if you divide 150/20 it will equal 7.5 then multiply 7.5 by 30 and you’ll get 225

7 0

3 years ago

400g of rapberries and 300g of strawberries cost £7.46. 500g of strawberries cost £4.10 work out the cost of 300g of raspberries

MAXImum [283]

Answer: £5.39

Step-by-step explanation:

500g strawberries = 4.10

100 g = 0.82

300g = 2.46

400g raspberries = 7.46 - 2.46 = 5.00

100g = 1.25

300g = 3.75

200g strawberries = 0.82 * 2 = 1.64

3.75 + 1.64 = 5.39

8 0

3 years ago

Find the area of the circle and explain how u got the answer . thanks

trasher [3.6K]

A = (pi)r^2

diameter = 12 in

radius = d/2 = 6 in

A = (pi)6^2

A = 36(pi)

3 0

3 years ago

Georgia works part-time at a daycare while she is going to college. She earned $160 last week. Georgia worked 12 more hours than

Lelu [443]

Answer:

B. $8

Step-by-step explanation:

Given:

Total payment, T = $160

Hourly rates = $t/hour

= total payment/total time worked

The question said, "worked 12 more hours than the amount she is paid per hour".

Total time worked = (12 + t) hour

Therefore, Hourly rates of payment, t = 160/(12 + t)

t^2 + 12t - 160 = 0

Solving the above quadratic equation,

t = -20 and 8

t = $8

3 0

3 years ago

The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha

Xelga [282]

Answer:

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

Step-by-step explanation:

Let suppose that function $f$ is continuous and integrable in the given intervals, by integral definition of average we have that:

$\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6$ (1)

$\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20$ (2)

By Fundamental Theorems of Calculus we expand both expressions:

$\frac{F(3)-F(-2)}{3-(-2)} = -6$

$F(3) - F(-2) = -30$ (1b)

$\frac{F(5)-F(3)}{5-3} = 20$

$F(5) - F(3) = 40$ (2b)

We obtain the average value of $f$ over the interval $[-2, 5]$ by algebraic handling:

$F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)$

$F(5) - F(-2) = 10$

$\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}$

$\bar f = \frac{10}{7}$

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

4 0

3 years ago