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3 years ago

Which situation could be described by the expression m-1.85

Mathematics

2 answers:

Bess [88]3 years ago

4 0

I think it whoud be 12 3

mezya [45]3 years ago

4 0

I think the answer is 12.3 if it is not right please reply back
Hope it Helps

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Your basketball team scored 4 fewer than twice as many points, x, as the other team. a. Write an expression for the number of po

velikii [3]

Answer:

A) p = 2x - 4 for the expression. B) Your team scored 44 points.

Step-by-step explanation:

Let <u>P</u> represent the points your team made.

Since the question says "Your basketball team scored 4 fewer than twice as many points as the other team." this is how I came up with the expression.

p = 2x - 4 ( our points = 2 times more then the other teams points minus 4.

Using this expression and the given information that the other team made 24 points, I solved B.

44 = 2(24) - 4.

3 0

3 years ago

Can you turn these inequalities into y=mx+b form? <br> y> 3x-6<br><br><br> 4x-y<7

kaheart [24]

I am not sure but wouldn't it be y=3x-6 and y=4x-7

8 0

3 years ago

Write an equation that models the sequence 400, 200, 100, 50, ...

Julli [10]

You're starting at 400. Each time you're cutting in half.

So the common ratio is 1/2 which means the equation is

$\Large y = 400(1/2)^{x-1}$

which is in the geometric sequence format. So the answer is choice D

3 0

3 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

suppose a study estimated that 78% of the residents of a town (with an error range plus/minus12 percentage points at 95% confide

abruzzese [7]

the Answer is 69% -Apex

4 0

3 years ago

Read 2 more answers