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3 years ago

ALEGBRA 1 ! help. me. please. this. is. so. hard. i dont. understand it.

Mathematics

1 answer:

ANEK [815]3 years ago

8 0

Answer:

Part a) to find the maximum height of the snowball, you have to differentiate the function. Therefore you get ----> dh/dt= -32x-8 . Now equate this to zero and solve for x. x= (-1)/4 now sub this value in to find h(x) [note: i'm talking about the original function] . I got h max = h((-1)/4) = 9 which is the max height.

Step-by-step explanation:

I'm not too sure about the other questions. Sorry

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If a photo measuring 4 inches by 6 inches is placed in a photo frame that is 2 inches wide all around, what percent of the frame

BARSIC [14]

It would be 2 by 2 in inches as the frame can only fit that much , which the area framed would be 4 inches as you multiply 2×2=4

8 0

3 years ago

Read 2 more answers

Which shows the graph of the solution set of 3y – 2x > –18?

nikdorinn [45]

Answer:

Graph of the inequality 3y-2x>-18 is given below.

Step-by-step explanation:

We are given the inequality, 3y-2x>-18

Now, using the 'Zero Test', which states that,

After substituting the point (0,0) in the inequality, if the result is true, then the solution region is towards the origin. If the result is false, then the solution region is away from the origin'.

So, after substituting (0,0) in 3y-2x>-18, we get,

3\times 0-2\times 0>-18

i.e. 0 > -18, which is true.

Thus, the solution region is towards the origin.

Hence, the graph of the inequality 3y-2x>-18 is given below.

8 0

3 years ago

Marcy bought 10.32 pounds of white potatoes and 12.48 pounds of sweet potatoes. About how many pounds of potatoes did Marcy buy?

zalisa [80]

Marcy bought a total of 22.8 potatoes. 10.32+12.48=22.8

3 0

3 years ago

Remember to show work and explain. Use the math font.

MrMuchimi

Answer:

$\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}$

Step-by-step explanation:

$\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================$

$1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}$

$\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)$

$--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)$

$--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}$

6 0

3 years ago

The area (in square feet) of the school sign can be represented by

yawa3891 [41]

Answer:

5x^2+2x-432

Step-by-step explanation:

15x^2+x-215×2+x−2

=15x^2+x-430+x−2

=15x^2+x+x-430-2

=15x^2+2x-432

3 0

2 years ago