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3 years ago

Please help! Could you answer all!?

Mathematics

2 answers:

Alexus [3.1K]3 years ago

8 0

My answer was wrong oops.

Brums [2.3K]3 years ago

3 0

Answer:

1.) 48

2.) 65

3.) 36

Step-by-step explanation:

1.) If the equation is 6(x-4) and x = 12, then all we have to do is plug in the value of x. When we plug in, all we do is substitute 12 for x because they mentioned in the question that x = 12. So, we end up getting 6(12 - 4). After solving this, we get 48.

2.) This problem is a lot like the last problem. All we need to do is substitute /plug in the values of x and y into the equation, to get 4(4^2) - 35/7 - (8 + 14). After solving, we get 65.

3.) . This problem, once again, is also a lot like the last problems. We need to substitute the value of x into the equation 8x+12. Since we know from the problem that x is 3, all we have to do is 8 * 3 + 12.

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The simplified expression

Romashka-Z-Leto [24]

Answer:

$5x^2 y^2$

Step-by-step explanation:

We need to use the properties shown below to solve this:

1. $\sqrt[n]{x^a} =x^{\frac{a}{n}}$

2. $\sqrt{x}\sqrt{x} =x$

3. $\sqrt{x} \sqrt{y}=\sqrt{x*y}$

Area of a triangle is given by 1/2 * base * height, so we do that and simplify:

$A=\frac{1}{2}(\sqrt{5x^3} )(2\sqrt{5xy^4} )\\A=\frac{1}{2}(5x^3)^{\frac{1}{2}}*2*(5xy^4)^{\frac{1}{2}}\\A=\sqrt{5}x^{\frac{3}{2}}*\sqrt{5}\sqrt{x} } y^2\\A=\sqrt{5} \sqrt{5}x^{\frac{3}{2}} x^{\frac{1}{2}}y^2\\A=5*x^2y^2\\A=5x^2 y^2$

6 0

3 years ago

At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

6 0

9 months ago

A bucket of paint has spilled on a tile floor. The paint flow can be expressed with the function p(t) = 6t, where t represents t

Neko [114]

Answer:

A) A[p(t)] = 36πt²

B) 7234.56 square units

Step-by-step explanation:

Given functions:

$\begin{cases}p(t)=6t \\ A(p)=\pi p^2 \end{cases}$

Part A

To find the area of the circle of spilled paint as a function of time, substitute the function p(t) into the given function A(p):

$\begin{aligned}A(p) & = \pi p^2\\\\ \implies A[p(t)] & = \pi [p(t)]^2\\& = \pi (6t)^2\\& = \pi 6^2 t^2\\& = 36\pi t^2\end{aligned}$

Part B

Given

t = 8 minutes
π = 3.14

Substitute the given values into the equation for A[p(t)} found in part A:

$\begin{aligned}A[p(8)] & = 36\pi t^2\\& = 36 \cdot 3.14 \cdot 8^2\\& = 36 \cdot 3.14 \cdot 64\\& = 113.04 \cdot 64\\& = 7234.56\:\: \sf square\:units\end{aligned}$