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2 years ago

Choose the correct slope of the line that passes through the points (−4, 8) and (−3, −6) ?

Mathematics

2 answers:

Andru [333]2 years ago

7 0

Answer:

-14

Step-by-step explanation:

$m=\frac{rise}{run} =\frac{-6-8}{-3+4}=\frac{-14}{1}=\boxed{-14}$

Hope this helps.

amm18122 years ago

5 0

Answer:

-14

Step-by-step explanation:

Slope = 8-(-6)/-4-(-3( = 8+6/-4+3 = 14/-1 = -14

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Fifteen cards numbered 1-15 are placed in a hat. What is the probability that the card has a multiple of 3 on it, given that the

Ivan

1-15 has only 1,3,5,7,9,11,13,15 as odd number therefore a total of 8 odd numbers within 1-15 so that would be 8/15 probability of it being an odd number (I’m not too sure of this sorry)

4 0

3 years ago

Translate each algebraic expression into words. a) x ÷ 12 b) 8m

CaHeK987 [17]

A.) a number divided by 12
b.) 8 times a number

3 0

3 years ago

Use the Distributive Property to express 24 + 40.

timurjin [86]

Answer:

4(6+10)

Step-by-step explanation:

4×6=24

4×10=40

5 0

3 years ago

Please help!!!!!!!!!!!!!!!!

vovikov84 [41]

i think that it is the third one

6 0

3 years ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

Zero Product Property: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the Zero Product Property, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$

Therefore, the solutions to the given equation are: x = 1 or x = -1

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3 0

1 year ago