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3 years ago

What’s the answer to this?

Mathematics

2 answers:

Galina-37 [17]3 years ago

6 0

Answer:

(x-8y)(x-8y)

Step-by-step explanation:

write the expression as a product with the factors x and 8y

8^2 -2(x)8y+64y^2

multiply terms with equal exponents by multiplying the bases

x^2 -2 (x) 8y + (8y)^2

factor

(x-8y)^2 which is equal to c

Salsk061 [2.6K]3 years ago

5 0

I just took this test too the answer is C.(x-8y)(x-8y)

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A radar is designed to report the track of an aircraft every 2 seconds. If this radar reports 15 reports 15 tracks in one minute

meriva

Answer:

Step-by-step explanation:

Since radar reports 30 tracks in one minute in which 60 seconds makes 1 minute

Let calculate using this formula

Using this formula

Percentage of time=Tracks reported in minutes/60 seconds*100

Let plug in the formula

Percentage of time=30 tracks/60 seconds*100

Percentage of time=0.5*100

Percentage of time=50%

Inconclusion The percentage of the time that radar track the aircraft is 50%

5 0

3 years ago

Read 2 more answers

2. A sum of $400 was shared between Samantha and Jade. Samantha received S100 more than Jade. a) If Jade received $150, how much

Marina CMI [18]

Answer:

Samantha share =$250

ratio = 5:3

Step-by-step explanation:

Samantha

$= jade + 100$

jade

$150$

samantha

$150 + 100 = 250$

ratio

250:150

5:3

5 0

2 years ago

What do ducks short necks help them to do? pls help :/

Kruka [31]

Answer:

In some circumstances these displays may allow the female to observe the performance of males and to evaluate them as potential mates. To elicit displays from a group of males, a female Mallard may swim with her neck outstretched and her head just above the water

Step-by-step explanation:

hope this helps

8 0

3 years ago

The height h (in feet) of an object dropped from a ledge after x seconds can be modeled by h(x)=−16x2+36 . The object is dropped

kakasveta [241]

Check the picture below.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&\\ \qquad \textit{at "t" seconds} \end{cases}$

so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?

$\bf h(x)=-16x^2+36\implies \stackrel{h(x)}{0}=-16x^2+36\implies 16x^2=36 \\\\\\ x^2=\cfrac{36}{16}\implies x^2 = \cfrac{9}{4}\implies x=\sqrt{\cfrac{9}{4}}\implies x=\cfrac{\sqrt{9}}{\sqrt{4}}\implies x = \cfrac{3}{2}~~\textit{seconds}$

now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.

$\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill$

quick info:

in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².

4 0

3 years ago

Bill and Henry were in a long jump competition. Bill went first and jumped 14.3 feet. If Henry jumped 1.3 times as far as Bill,

emmainna [20.7K]

B = 14.3
H = 1.3B

H = 1.3(14.3) = 18.59 ft <==

3 0

3 years ago

Read 2 more answers