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3 years ago

Need answers to three questions in picture will give brainly

Mathematics

2 answers:

masya89 [10]3 years ago

7 0

Answer:

3. B

4. J

5. C

Trava [24]3 years ago

6 0

Answer:

3. the answer is A3

Step-by-step explanation:

you do 3x-1 and you get 3

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Cylinder A has a radius of 10 inches and a height of 5 inches. Cylinder B has a volume of 750π. What is the percentage change in

Firlakuza [10]

Answer:

50% change in volume

Step-by-step explanation:

<h2>This problem bothers on the mensuration of solid shapes.</h2>

In this problem we are to find the volume of the first cylinder and compare with the second cylinder.

Given data

Volume v = ?

Diameter d= ?

Radius r = $10 in$

Height h= $5 in$

we know that the volume of a cylinder is expressed as

$volume = \pi r^{2}h$

Substituting our given data we have

$volume = \pi*10^{2}*5\\ volume= \pi *100*5\\volume= 500\pi in^{3} \\$

The first cylinder as a volume of $500\pi$

The change in volume is $750\pi - 500\pi = 250\pi$

percentage = $\frac{250\pi }{500\pi } *100$

$percentage=$ $0.5*100= 50$ %

6 0

3 years ago

Read 2 more answers

Garrett chooses to invest $1800 in a simple. The account earns 3% interest over 5 years which equation represents garrets invest

Alla [95]

The interest earned in 5 years would be $270

<u>Explanation:</u>

Given:

Principal, P = $1800

Rate of interest, r = 3%

Time, t = 5 years

Simple interest, I = ?

We know,

$I = \frac{p X r X t}{100}$

On substituting the value we get

$I = \frac{1800 X 3 X 5}{100} \\\\I = 270$

Therefore, interest earned in 5 years would be $270

6 0

4 years ago

In how many ways may one A, three B’s, two C’s, and one F be distributed among seven

astraxan [27]

7 different ways is the answer

5 0

3 years ago

Root 3 cosec140° - sec140°=4<br>prove that<br><br>

Lorico [155]

Answer:

Step-by-step explanation:

We are to show that $\sqrt{3} cosec140^{0} - sec140^{0} = 4\\$

<u>Proof:</u>

From trigonometry identity;

$cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}$

$\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\$

From trigonometry, 2sinAcosA = Sin2A

$= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}$