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2 years ago

Please help lol, im on a math test

Mathematics

1 answer:

Sveta_85 [38]2 years ago

5 0

I know for sure it’s not B or D so I believe it is A.

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The variables X, Y, and Z represent polynomials where X = a, Y = 3a – 5, and Z = a^2 + 2. What is X^2Y – Z in simplest form?

klasskru [66]

$X=a \\
Y=3a-5 \\
Z=a^2+2 \\ \\
X^2Y-Z=a^2(3a-5)-(a^2+2)=3a^3-5a^2-a^2-2=3a^3-6a^2-2$

The answer is b.

4 0

3 years ago

Read 2 more answers

Heeeelp me plz i will give brainliest

katen-ka-za [31]

Most of the time, plants are producers because they get their food from photosynthesis. Lily pads do, so it is a producer.

Hope this helps!

3 0

3 years ago

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Find the percent of decrease if 110 is decreased to 88 A) 125% B)80% C)20%

stich3 [128]

Hey there! I'm happy to help!

We have 110 minus a certain percent of 110 equals 88. Let's call this percent p and solve.

110-110p=88

We subtract 110 from both sides.

-110p=-22

We divide both sides by 110.

p=0.2

Therefore, the percent decrease is C) 20%.

Have a wonderful day! :D

4 0

2 years ago

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Ashley wants to pour 15.82 grams of salt into a container. So far, she has poured 13.87 grams. How much more salt should Ashley

erik [133]

Answer:

1.95

Step-by-step explanation:

Basically, you subtract: 5.82 - 13.87 = 1.95

Then, see: 15.82 - 1.950 = 13.87

Then you have your answer :D

Ashley has to pour 1.95 more grams of salt.

Brainliest, please??? :>

3 0

2 years ago

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Evaluate the limit of sequence:

mr_godi [17]

Rationalize both the numerator and denominator. Given

$\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d}$

we can rationalize it by introducing conjugates of the numerator and denominator:

$\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d} \cdot \dfrac{\sqrt a+\sqrt b}{\sqrt a+\sqrt b} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt c+\sqrt d} \\\\ = \dfrac{\left(\sqrt a\right)^2 - \left(\sqrt b\right)^2}{\left(\sqrt c\right)^2-\left(\sqrt d\right)^2} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b} \\\\ = \dfrac{a-b}{c-d} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b}$

Then the limit is equivalent to

$\displaystyle \lim_{n\to\infty} \frac{(n+3)-n}{(n+1)-n} \cdot \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n} = 3 \lim_{n\to\infty} \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n}$

For the remaining expression, divide through uniformly by $\sqrt n$ :

$\dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n} = \dfrac{\sqrt{1+\frac1n} + 1}{\sqrt{1+\frac3n}+1}$

As <em>n</em> goes to infinity, the remaining terms containing <em>n</em> converge to 0, leaving

$\dfrac{\sqrt{1}+1}{\sqrt1+1} = \dfrac22 = 1$

making the overall limit 3.

8 0

2 years ago