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dalvyx [7]
3 years ago
11

Please answer ive been having such a bad day today

Mathematics
1 answer:
madreJ [45]3 years ago
5 0

Answer:

A - y = 2x + 1

Step-by-step explanation:

2(2) + 1 = 5

2(4) + 1 = 9

2(6) + 1 = 13

Hope this helps <3

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At a restaurant a meal used to cost $9, but a new owner decided to raise the price by 13
kari74 [83]

Answer:

$10.17

Step-by-step explanation:

9 x .13 = 1.17

9 + 1.17 = 10.17

6 0
1 year ago
What's the equation for this problem? Please help!! It's due tomarrow
prohojiy [21]
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3 years ago
Geometry question need help finding the answer.
Xelga [282]

Answer: 1208.96

Step-by-step explanation:

The volume of the cylinder is (\pi)(4^2)(4)=64\pi

The volume of the prism is (6)(12)(14)=1008

So, the total volume is 1008+64\pi=\boxed{1208.96}

6 0
2 years ago
Erik and Caleb were trying to solve the equation: 0=(3x+2)(x-4) Erik said, "The right-hand side is factored, so I'll use the zer
Mumz [18]

Answer:

C) Both

Step-by-step explanation:

The given equation is:

0=(3x+2)(x-4)

To solve the given equation, we can use the Zero Product Property according to which if the product <em>A.B = 0</em>, then either A = 0 OR B = 0.

Using this property:

(3x+2) = 0 \Rightarrow \bold{x = -\frac{2}{3}}\\(x-4) = 0 \Rightarrow \bold{x = 4}

So, Erik's solution strategy would work.

Now, let us discuss about Caleb's solution strategy:

Multiply (3x+2)(x-4) i.e. 3x^2-12x+2x-8 = 3x^2-10x-8

So, the equation becomes:

0=3x^2-10x-8

Comparing this equation to standard quadratic equation:

ax^2+bx+c=0

a = 3, b = -10, c = -8

So, this can be solved using the quadratic formula.

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{-(-10)\pm\sqrt{(-10)^2-4\times3 \times (-8)}}{2\times 3}\\x=\dfrac{-(-10)\pm\sqrt{196}}{6}\\x=\dfrac{10\pm14}{6} \\\Rightarrow x= 4, -\dfrac{2}{3}

The answer is same from both the approaches.

So, the correct answer is:

C) Both

3 0
3 years ago
If f(x) = |x| + 9 and g(x) = –6, which describes the range of (f + g)(x)?
Leokris [45]
Just want points byeeee hope u fail jkjkjk
3 0
3 years ago
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