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mihalych1998 [28]
3 years ago
15

For which equation is y = 7 a solution?

Mathematics
2 answers:
JulsSmile [24]3 years ago
6 0
B

Because when you have 18=11+y you have to subtract 11 from the other side which will give you 7=y which is the same as y=7
natali 33 [55]3 years ago
3 0

Answer:

B

Step-by-step explanation:

11+7=18

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Ƒ(x) = x2 + 3x - 10. Determine the zeros of the equation.
7nadin3 [17]
I am assuming the equation is as follows
f(x) = x^2 + 3x - 10

Since the constant 10 is -10 this tells us we are going to have different signs + - after you have factored the equation.

First thing first. Think of two numbers that will give you the product(The answer after two numbers are multiplied) of -10 and the sum(Two numbers added together) of +3. I get the 3 from the 3x.

We know 5 x -2 = -10 and the sum 5 + -2 =  +3
With that said, we found our two factors
(x + 5)(x - 2)

Now write them as zeros and solve for x
x + 5 = 0
x = -5

x - 2 = 0
x = 2

So the zeros are -5 and 2.


3 0
4 years ago
Lisa specializes in baking lemon cupcakes. She bakes 3 dozen cupcakes every hour. The cost (in dollars) of making n cupcakes is
kirza4 [7]

Answer:

36h, 60+16.2h, and $92.40.

Step-by-step explanation:

n(h)=36h because there are 3 dozen cupcakes made per hour, to get C(h) you just plug in 36h in place of n and simplify, and then for the final answer 2 is plugged into the equation from the second answer.

8 0
4 years ago
I need this by tomorrow<br>​
erastova [34]

Answer:

r*4.50+5=t

Step-by-step explanation:

8 0
4 years ago
Algebra 2A Help !<br><br> Tadakimasu ~
Ludmilka [50]
Function is p(x)=(x-4)^5(x^2-16)(x^2-5x+4)(x^3-64)

first factor into (x-r1)(x-r2)... form

p(x)=(x-4)^5(x-4)(x+4)(x-4)(x-1)(x-4)(x^2+4x+16)
group the like ones
p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)

multiplicity is how many times the root repeats in the function
for a root r₁, the root r₁ multiplicity 1 would be (x-r₁)^1, multility 2 would be (x-r₁)^2 

so

p(x)=(x-4)^8(x+4)^1(x-1)^1(x^2+4x+16)
(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the  root 1 and has multiplity 1
(x^2+4x+16) is not on the real plane, but the roots are -2+2i√3 and -2-2i√3, each multiplicity 1 (but don't count them because they aren't real

baseically

(x-4)^8 is the root 4, it has multiplicity 8
(x-(-4))^1 is the root -4 and has multiplicity 1
(x-1)^1 is the  root 1 and has multiplity 1

7 0
3 years ago
What's the answerrrrrrrrrrrrrr
Evgesh-ka [11]
I tnjnk is b I'm not sure
5 0
3 years ago
Read 2 more answers
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