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3 years ago

I need help with this ' ASAP!

Mathematics

2 answers:

pishuonlain [190]3 years ago

7 0

The answer is 120 square units. The area of the shaded units combined is 120 and the area of the entire shape is 240. So when you minus 120 from 240, then you get 120 left as your answer.

Taya2010 [7]3 years ago

4 0

Answer:

H. 120

Step-by-step explanation:

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Fatima’s softball practice starts at 9:30 a.m. Practice lasts for 1 hour and 10 minutes.

ki77a [65]

10:40 am.

9:30 plus 1 hour = 10:30 plus ten minutes= 10:40

3 0

3 years ago

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Center: (-2,5), radius:7

Firlakuza [10]

Answer:

(x + 2)² + (y - 5)² = 7²

Step-by-step explanation:

Incomplete question. I believe you meant, "write the equation of the circle with center at (-2, 5) and radius 7.

That would be:

(x + 2)² + (y - 5)² = 7²

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3 years ago

Please show the work as well 5TH GRADE MATH

Liono4ka [1.6K]

Answer:

$9450 {m}^{3}$

Step-by-step explanation:

Given ,

Length of the tank = 27m

Width of the tank = 35m

Depth of the tank = 10m

Therefore, 

Volume of the tank = Length × Width × Depth

= 27m × 35m × 10m

$= 9450 {m}^{3} (ans)$

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3 years ago

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Need help with 5 6 7

sineoko [7]

5) 1,512 is 5 because 12,600× .12 leaves you 1512$ dollars off the original price.

6) 15.38 or 15.4% if you round to the nearest tenth. 32,800÷ 2132 leaves you the % additional of the raise can also be written as 32,800 = 2132x divide both sides and x being the percent is 15.4%.

7) the discount is 49.90$ another one like 5. multiply 499.00 x .10 = 49.9 also said as 49.90.

8 0

3 years ago

P = √mx_ £x MX t make x the subject

vladimir2022 [97]

So, making x subject of the formula, x = [m - 2pt³ ±√(m² - 4pt²m)]/{2t⁵}

<h3>How to make x subject of the formula?</h3>

Since p = √(mx/t) - t²x

So, p + t²x = √(mx/t)

Squaring both sides, we have

(p + t²x)² = [√(mx/t)]²

p² + 2pt²x + t⁴x² = mx/t

Multiplying through by t,we have

(p² + 2pt²x + t⁴x²)t = mx/t × t

p²t + 2pt³x + t⁵x² = mx

p²t + 2pt³x + t⁵x² - mx = 0

t⁵x² + 2pt³x - mx + p²t = 0

t⁵x² + (2pt³ - m)x + p²t = 0

Using the quadratic formula, we find x.

$x = \frac{-b +/-\sqrt{b^{2} - 4ac} }{2a}$

where

a = t⁵,
b = (2pt³ - m) and
c = p²t

Substituting the values of the variables into the equation, we have

$x = \frac{-(2pt^{3} - m) +/-\sqrt{(2pt^{3} - m)^{2} - 4(t^{5})(p^{2}t) } }{2t^{5} }\\= \frac{-(2pt^{3} - m) +/-\sqrt{4p^{2} t^{6} - 4pt^{2}m + m^{2} - 4p^{2}t^{6} } }{2t^{5}}\\= \frac{-(2pt^{3} - m) +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}\\= \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}$

So, making x subject of the formula, $x = \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}$

Learn more about subject of formula here:

brainly.com/question/25334090

#SPJ1

6 0

1 year ago