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baherus [9]
3 years ago
7

On a science test, 2 points are deducted from a total of 100 points for each question answered incorrectly. Trevor scored an 84

on the test. The equation 100 – 2x = 84, where x represents the number of questions answered incorrectly, represents the situation. How many questions did Trevor answer incorrectly?
questions
Mathematics
2 answers:
ale4655 [162]3 years ago
6 0
He answered 8 questions incorrectly. 2*8=16 100-16=84
sattari [20]3 years ago
3 0
The answer is 8     i have to have 20 characters

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Which expression is equivalent to 64x^3-343
In-s [12.5K]

Answer:

B:  (4x - 7)(16x^2 + 28x + 49)

Step-by-step explanation:

64x^3-343 can be rewritten as 4^3x^3 - 7^3 or (4x)^3 - 7^3.  This is the difference of two cubes.  The appropriate formula for factoring such is

       a^3 - b^3 = (a - b)(a^2 + ab + b^2).  Therefore,

our (4x)^3 - 7^3 = (4x - 7)(16x^2 + 28x + 49).  This is Answer B.

4 0
3 years ago
Pls help me
Mashutka [201]

Answer:

A is the answer because...

Step-by-step explanation:

Let's narrow it down, so it is not D because 8/2=4 but 64/4=16, so already they don't share a slope. It is not B because the numbers don't work together like they should, since we are trying to prove proportionality. It is not C because we are not trying to square the numbers but find a similarity between the numbers, the same slope I mean. So that leaves A which has  the slope of 2, test it out now. 0 and 0 are the same, 4/2=2, 8/4=2, 12/6=2. So now you know that A is the correct answer.

Hope you understand! :)

3 0
3 years ago
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kiruha [24]

answer:

4(2+x) and 8 + 4x

since she memorized those scales for EACH of those four lessons, the answer would be that.

7 0
3 years ago
Question regarding logarithms.
Eddi Din [679]

5^{x-2}-7^{x-3}=7^{x-5}+11\cdot5^{x-4}\\\\5^{x-2}-7^{x-2-1}=7^{x-2-3}+11\cdot5^{x-2-2}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\5^{x-2}-\dfrac{7^{x-2}}{7^1}=\dfrac{7^{x-2}}{7^3}+11\cdot\dfrac{5^{x-2}}{5^2}\\\\5^{x-2}-\dfrac{1}{7}\cdot7^{x-2}=\dfrac{1}{343}\cdot7^{x-2}+\dfrac{11}{25}\cdot5^{x-2}\\\\-\dfrac{1}{7}\cdot7^{x-2}-\dfrac{1}{343}\cdot7^{x-2}=\dfrac{11}{25}\cdot5^{x-2}-5^{x-2}\\\\\left(-\dfrac{1}{7}-\dfrac{1}{343}\right)\cdot7^{x-2}=\left(\dfrac{11}{25}-1\right)\cdot5^{x-2}

\left(-\dfrac{49}{343}-\dfrac{1}{343}\right)\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\\\\-\dfrac{50}{343}\cdot7^{x-2}=-\dfrac{14}{25}\cdot5^{x-2}\qquad\text{multiply both sides by}\ \left(-\dfrac{25}{14}\right)\\\\\dfrac{50\cdot25}{343\cdot14}\cdot7^{x-2}=5^{x-2}\qquad\text{divide both sides by}\ 7^{x-2}\\\\\dfrac{25\cdot25}{343\cdot7}=\dfrac{5^{x-2}}{7^{x-2}}\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}

\dfrac{5^2\cdot5^2}{7^3\cdot7}=\left(\dfrac{5}{7}\right)^{x-2}\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{5^4}{7^4}=\left(\dfrac{5}{7}\right)^{x-2}\\\\\left(\dfrac{5}{7}\right)^4=\left(\dfrac{5}{7}\right)^{x-2}\iff x-2=4\qquad\text{add 2 to both sides}\\\\\boxed{x=6}

6 0
3 years ago
Read 2 more answers
Consider the following vector-valued function:~h(t) =〈2 sin(3t),3 cos(3t),√5 sin(3t)〉0≤t≤2π3This defines a smooth parameterized
omeli [17]

Answer:

a) h'(t)= (6cos3t,-9sin3t,3\sqrt[]{5}cos3t)

b) t=0.93994736+πn/3

c) Magnitude of h(t) is 3 which is a constant, so h(t) lies on a sphere

Step-by-step explanation:

5 0
3 years ago
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