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klasskru [66]
3 years ago
15

Jackie bicycles 9 kilometers west to get from her house to school. After school, she bicycles 12 kilometers north to her friend

Ronald's house. How far is Jackie's house from Ronald's house, measured in a straight line?
Mathematics
1 answer:
klemol [59]3 years ago
8 0
Answer:

21 kilometers!

Explanation: You add the two distances she rode her bicycle and add them together. 9+12 = 21kilometers!
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Need help with this??? Please!!!
Radda [10]

Answer:

the first option

Step-by-step explanation:

(f-g)(x) simply means to subtract both expressions. really literally.

and we go through it power by power of x.

the highest power/exponent of x is 3 (x³). only f(x) has one.

so, -7x³ is the first part of f-g.

next is x².

11x² - 6x² = 5x², which is the second part of f-g.

next is x.

-8x - (-14x) = -8x + 14x = 6x, which is the third part of f-g.

next is x⁰ (in other words, no x, just a constant).

4 - (-3) = 4 + 3 = 7, which is the 4th part of f-g.

we have no x to the power of -1 or -2, so we have -3

0 - (-4x‐³) = 4x‐³, which is the last part of f-g.

so, it is clearly the first answer option.

3 0
3 years ago
The graph of f(x) = x2 – 3x2 + 4 is shown
IRISSAK [1]

Answer:

2 is the answer that I would give. The question says <u>distinct</u> solutions does this graph have.

Step-by-step explanation:

You are asked only how many times this crosses the x axis. You are not asked for an exact value. So the thing to do is graph the polynomial.

y = x^3 - 3x^2 + 4

The graph says that x = -1 with a multiplicity of 1

It also says that x=2 is a solution with a multiplicity of 2 (meaning there are 2 roots are the same).

So this polynomial factors into y = (x + 1)(x - 2)(x - 2)

3 0
3 years ago
The sum of three consecutive terms in an arithmetic sequence is 27, and their product is 585. find the three terms.
sammy [17]
\text{the three consecutive terms in an arithmetic sequence}\\a;\ a+d;\ a+2d\\\\\text{system of equals}\\\\ \left\{\begin{array}{ccc}a+a+d+a+2d=27\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}3a+3d=27&|:3\\a(a+d)(a+2d)=585\end{array}\right\\ \left\{\begin{array}{ccc}a+d=9&\to d=9-a\\a(a+d)(a+2d)=585\end{array}\right\\
\text{substitute to the second equation}\\\\a(a+9-a)(a+2(9-a))=585\\\\a(9)(a+18-2a)=585\\9a(18-a)=585\ \ \ |:9\\a(18-a)=65\\18a-a^2=65\ \ \ |\text{change the signs}\\a^2-18a=-65\\a^2-2a\cdot9=-65\ \ \ |+9^2\\\underbrace{a^2-2a\cdot9+9^2}_{(a-b)^2=a^2-2ab+b^2}=-65+9^2\\(a-9)^2=16\to a-9\pm\sqrt{16}\\a-9=-4\ \vee\ a-9=4\ \ \ |+9\\a=5\ \vee\ a=13\\\\d=9-5=4\ \vee\ d=9-13=-4
Answer:\ a=5;\ a+d=9; a+2d=13\ vee\ a=13;\ a+d=9;\ a+2d=5

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Answer:

1.) D

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4 0
4 years ago
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