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4 years ago

Help pls image attached

Mathematics

2 answers:

zimovet [89]4 years ago

6 0

Answer:

Step-by-step explanation:

lapo4ka [179]4 years ago

3 0

Answer:

Last Option (8,6) is the right answer.

Step-by-step explanation:

The given function is f(x) = 8 - ∛x

We can rewrite the function f(x) = 2³ - ∛x as y = 2³ - ∛x

Now we will check for each point whether it lies on the graph.

For (-64,12)

y = 8 - ∛-64

y = 8 - ∛(-4)³ = 8 - (-4) = 12

Therefore this point lies on the graph.

For point (125,3)

y = 8 - ∛125

y = 8 -5 =3

Therefore the point lies on this graph.

For point (343,1)

y = 8 - ∛343

y = 8 - ∛7³ = 8 - 7 = 1

Therefore the point lies on graph.

For point (8,5)

y = 8 - ∛8

y = 8 - 2 = 6 ≠ 5

While the value of x coordinate of the given point is 8.

Therefore this point doesn't lie on the given graph.

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Simplify 25^ - 1/2 <br><br>Please explain! I know the answer is 1/5 but why?

lisov135 [29]

$(25)^{-\frac{1}{2} }$

When the exponent is negative, you have to make the number a fraction(1 over the number).

For example:

$25^{-2} = \frac{1}{25^{2} }$

$x^{-1} =\frac{1}{x^{1} }$

This gets rid of the negative, but it leaves the exponent

$(25)^{-\frac{1}{2}} = \frac{1}{25^{1/2}}$

When an exponent is a fraction, it is the square root.

For example:

$x^{\frac{1}{3}} = \sqrt[3]{x}$

$x^{\frac{1}{2}} =\sqrt{x}$

$x^{\frac{1}{4}}=\sqrt[4]{x}$

So...

$\frac{1}{25^{1/2}} = \frac{1}{\sqrt{25} } = \frac{1}{5}$

3 0

4 years ago

Seven out of twenty giraffes like mince pies. Write this as a percent.

Tcecarenko [31]

.35 is the answer because when you write it as a fraction
7/20we know that 20 equals 100we get 100by multiplying 20 by 5 do you just multiply 7by5 which equals 35 over 100 which equals .35.please follow me on brainly

4 0

4 years ago

I need help in this please! thank you so much!

Anna35 [415]

Answer:

C) p(x) has a zero between x=2 and x=4

Step-by-step explanation:

For a polynomial p(x), $p(a)=R$ means that, when p(x) is divided by $x-a$ , the remainder is $R$ .

This is called the remainder theorem

Therefore $p(2)=1$ , $p(4)=-1$ means that when p(x) is divided $x-2$ and $x-4$ is 1 and $-1$ respectively.

Also the Intermediate value theorem says that, if p(a) and p(b) have opposite signs then there is a certain $c$ on $(a,b)$ where $f(c)=0$ i.e there is a zero between $a$ and $b$