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kodGreya [7K]
3 years ago
5

Find the prime factorization of each number. 1. 50 =

Mathematics
1 answer:
Varvara68 [4.7K]3 years ago
7 0

Answer:

1.5 is not a prime number

Step-by-step explanation:

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Chase purchased a sweatshirt from the clearance rack. The price of the sweatshirt after the discount is represented by the expre
Black_prince [1.1K]

Answer:B(0.75+0.25)x

Step-by-step explanation:

Is a discount of 0.75 was given the d real value of the sweatshirt would be a whole number

Which is 1-0.75=0.25

6 0
3 years ago
For their wedding, Rajai and Carly received $1000. Their financial advisor laid out 4 different options for them to invest in. A
dolphi86 [110]

Answer:

Their best investment when they retire in 40 years would be option B.

Step-by-step explanation:

Ragai and Carly invest the $1000 received for their wedding for 40 years.

From the diagram,

In option A, the initial investment do not increase at a constant rate yearly.

In option B, the amount invested increase by $75 yearly.

In option C, the yearly increase does not have a steady value.

In option D, the amount invested increases by a n + consecutive odd values yearly. Where n is the increase of the previous year.

Their best investment when they retire in 40 years would be option B because it would yield the highest profit.

8 0
3 years ago
Someone please help me with this lol… have no idea what I’m doing
Sholpan [36]

Given:

\cos \theta =\dfrac{3}{5}

\sin \theta

To find:

The quadrant of the terminal side of \theta and find the value of \sin\theta.

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

\cos \theta =\dfrac{3}{5}

\sin \theta

Here cos is positive and sine is negative. So, \theta must be lies in Quadrant IV.

We know that,

\sin^2\theta +\cos^2\theta =1

\sin^2\theta=1-\cos^2\theta

\sin \theta=\pm \sqrt{1-\cos^2\theta}

It is only negative because \theta lies in Quadrant IV. So,

\sin \theta=-\sqrt{1-\cos^2\theta}

After substituting \cos \theta =\dfrac{3}{5}, we get

\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}

\sin \theta=-\sqrt{1-\dfrac{9}{25}}

\sin \theta=-\sqrt{\dfrac{25-9}{25}}

\sin \theta=-\sqrt{\dfrac{16}{25}}

\sin \theta=-\dfrac{4}{5}

Therefore, the correct option is B.

6 0
2 years ago
The question is down bellow!!!
BartSMP [9]
D............a raisin is about 1 cm long

6 0
3 years ago
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lisov135 [29]

Answer:

A for Circle=2pi*r and A for Square= b*h, then find the difference between the two

Step-by-step explanation:

7 0
2 years ago
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