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3 years ago

Which expression shows a way to find 356+495?

Mathematics

2 answers:

kiruha [24]3 years ago

7 0

700 + 140 + 11 = it is more of expanded form and would result and the same answer. Hope this helps!

Yakvenalex [24]3 years ago

6 0

C. 700+140+11. You basically add the hundreds, tens, and ones.

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The prime factors of a number are 2, 2, 2, and 5. What is the number?

fgiga [73]

Answer: 40

Step-by-step explanation:

Since 40 could be = 1 x 40, 2 x 20, 4 x 10, or 5 x 8.

The Factors of 40 will therefore be: 1, 2, 4, 5, 8, 10, 20, 40.

The Prime factor then is: 2, 2, 2, 5

Multiplying them together gives 40.

7 0

3 years ago

Please help me!!!!! I don't want to get a bad grade and CAN YOU PLEASE SHOW YOUR WORK!!!!

Veseljchak [2.6K]

E. 53
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4 0

3 years ago

Read 2 more answers

Please help me out with this

Anettt [7]

Answer:

81.75 ft²

Step-by-step explanation:

The area (A) of a trapezoid is calculated using the formula

A = $\frac{1}{2}$ h (a + b)

where a, b are the parallel bases and h is the perpendicular height

Calculate h using the right triangle and the sine ratio

sin30° = $\frac{opposite }{hypotenuse}$ = $\frac{h}{10}$

Multiply both sides by 10

10 × sin30° = h, thus

h = 5

a = 12 and b = 8.7 + 12 = 20.7, hence

A = $\frac{1}{2}$ × 5 × (12 + 20.7)

= 0.5 × 5 ×32.7

= 81.75 ft²

7 0

3 years ago

What in 40 023 032 in word form??

krok68 [10]

Forty million, twenty-three thousand and thirty-two

5 0

3 years ago

In this exercise, consider a particle moving on a circular path of radius b described by r(t) = b cos(ωt)i + b sin(ωt)j, where ω

Bingel [31]

Answer:

Acceleration of the particle = $bw^{2}$

Step-by-step explanation:

We are given the position vector of a particle moving in a circle of radius b units.

r(t) = b cos(ωt)i + b sin(ωt)j

Velocity , v = $\frac{dr}{dt}$ = -bω sin(ωt)i + bω cos(ωt)j

The magnitude of velocity, v = $\sqrt{v_x^{2} +v_y^{2} }$

Squaring both sides,

$v^{2} = b^{2} w^{2}(sin^{2}(wt)+cos^{2}(wt))$

Since $sin^{2}(wt)+cos^{2}(wt))$ = 1

$v^{2} = b^{2}w^{2}$

The acceleration towards the centre is called the centripetal acceleration and is given by

a = $\frac{v^{2} }{r}$

a = $\frac{b^{2}w^{2}}{b}$

a = $bw^{2}$

3 0

3 years ago