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2 years ago

I need help with this question. Thanks! Also, the options is can or cannot

Mathematics

1 answer:

bulgar [2K]2 years ago

4 0

I can’t see the options lol show them

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Answer questions a) and b)

Mariulka [41]

Answer:

$x\hookrightarrow -1\hookrightarrow 0\hookrightarrow 1\hookrightarrow 2\hookrightarrow 3\hookrightarrow 4\hookrightarrow 5$

$y\hookrightarrow 5\hookrightarrow 0\hookrightarrow -3\hookrightarrow -4\hookrightarrow -3\hookrightarrow 0\hookrightarrow 5$

The curve (B) matches the graph of y=x²-4x.

$----------\\Hope \ it\;helps$

4 0

2 years ago

−t=9(t−10) what is t

Luda [366]

Answer:

t = 9

Step-by-step explanation:

First, distribute the 9.

Next, add subtract 9t from both sides.

Divide both sides by -10. When you divide a negative by another negative, they cancel each other out.

7 0

2 years ago

HELP ASAP 30 POINTS! GIVING BRAINLIEST TO THE CORRECT ANSWER

satela [25.4K]

Answer:

C. (x-8)^2 = 52 is the correct answer.

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

3 years ago

You toss two coins. what is the probability p that both are tails

Artist 52 [7]

25% or 1/4

1/2 × 1/2 = 1/4

3 0

2 years ago