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2 years ago

I need help please and explain your answer it’s number 4

Mathematics

1 answer:

Brums [2.3K]2 years ago

8 0

Answer:

Step-by-step explanation:

Well "value" of 3x^2 - 7x + 9 when x = 4 is just x plugged into the expression

3 * 4^2 - 7 * 4 + 9

3 * 16 - 28 + 9

= 29

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Write a multiplication problem in which the product has two decimal places

sukhopar [10]

.5 x .5 = .25

Here you go! :)

8 0

3 years ago

Read 2 more answers

For a brainlist +10 points please help

Strike441 [17]

Answer:

C. Two smaller atoms come together to form a larger atom that is a different element.

Explanation:

Fusion is a reaction in which two nuclei combine to form a nucleus with the release of energy.

8 0

2 years ago

If a=under root s(s-a)(s-b)(s-c) then the value of a, when a =3 b =4c = 5 and s =a+b+c by 2

kramer

Answer:

Step-by-step explanation:

Here it's given that ,

$\sf\red{\longrightarrow} A=\sqrt{s(s-a)(s-b)(s-c)}$

Also ,

$\sf\red{\longrightarrow} s =\dfrac{a+b+c}{2}$

And we need to find out the value of A , when

a = 3
b = 4
c = 5

So , on substituting the respective values to find s we have ,

$\sf\red{\longrightarrow} s =\dfrac{3+4+5}{2}=\dfrac{12}{2}=\bf{6}$

Now let's find out A as ,

$\sf\red{\longrightarrow} A = \sqrt{6(6-3)(6-4)(6-5)} \\$

$\sf\red{\longrightarrow}A =\sqrt{6 * 3 * 2 * 1}\\$

$\sf\red{\longrightarrow} A = \sqrt{3^2 * 2^2 *1^2}\\$

$\sf\red{\longrightarrow}A = 3*2\\$

$\sf\red{\longrightarrow}\boxed{\qquad \blue{\bf A = 6 \qquad}}$

7 0

2 years ago

When a linear function has a slope of 5, what is the "run" part of the slope?

alukav5142 [94]

Slope equals rise over the run
so if slope = 5 then the rise also equals 5 and the run is 1

7 0

3 years ago

Toby needs to save a minimum of $250. If Toby can save $57 a month, how many months will it take for him to save enough? Select

kiruha [24]

Answer:

$250 \leqslant 57m$

$57m \: \geqslant 250$

Step-by-step explanation:

Minimum is a word meaning greater than or equal to.

Hope this helps!

If you have anyother question feel free to ask!

7 0

1 year ago