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lord [1]
3 years ago
9

Ms. Turner drove 825 miles in March. She drove 3 times as many miles in March as she did in January. She drove 4 times as many m

iles in February as she did in January. What was the total number of miles Ms. Turner drove in February?
Mathematics
1 answer:
navik [9.2K]3 years ago
4 0
825/3 = 275 which is how many miles she drove in January.

275*4= 1100 which is how many miles she drove in February. 

Ms. Turner drove 1100 miles in February. 
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Use vieta's theorem to solve the problems.
Mnenie [13.5K]

Using Vieta's Theorem, it is found that c = 72.

<h3>What is the Vieta Theorem?</h3>
  • Suppose we have a quadratic equation, in the following format:

y = ax^2 + bx + c

  • The roots are p and q.

The Theorem states that:

p + q = -\frac{b}{a}

pq = \frac{c}{a}

In this problem, the polynomial is:

x^2 - 17x + c

Hence the coefficients are a = 1, b = -17.

Since the difference of the solutions is 1, we have that:

p - q = 1

p = 1 + q

Then, from the first equation of the Theorem:

p + q = -\frac{b}{a}

1 + q + q = 17

2q = 16

q = 8

p = 1 + q = 9

Now, from the second equation:

pq = \frac{c}{a}

72 = c

c = 72

To learn more about Vieta's Theorem, you can take a look at brainly.com/question/23509978

6 0
2 years ago
Plzzzzzzzzzzzzzzzzzzzzzz neeed help dew nowwwwwwwwwwwwwwwwwwwwwwwwwww
vampirchik [111]

Answer:

64 teachers

Step-by-step explanation:

We can use ratios to solve

14 students   896 students

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Using cross products

14 * x = 896*1

Divide each side by 14

14x/14 = 896/14

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4 0
3 years ago
A middle school has 600 students and 34% of them are in the 8th grade. One-fourth of 8th graders walk to school. How many 8th gr
Kaylis [27]

600 * .34 =  204 are in 8th grade

204 * 1/4 = 51 51 8th graders walk to school

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5 0
2 years ago
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Lim x approaches 0 (1+2x)3/sinx
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Interpreting your expression as

\dfrac{3(1+2x)}{\sin(x)}

when x approaches zero, the numerator approaches 3:

3(1+2x) \to 3(1+2\cdot 0) = 3(1+0) = 3\cdot 1 = 3

The denominator approaches 0, because \sin(0)=0

Moreover, we have

\displaystyle \lim_{x\to 0^-} \sin(x) = 0^-,\quad \displaystyle \lim_{x\to 0^+} \sin(x) = 0^+

So, the limit does not exist, because left and right limits are different:

\displaystyle \lim_{x\to 0^-} \dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^-} = -\infty,\quad \displaystyle \lim_{x\to 0^+}\dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^+} = +\infty

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V2 + 2v + 1 is the product of which two binomials?
juin [17]
(V+1)2 is that right
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