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3 years ago

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Brooklyn and Nicole are both driving along the same highway in two different cars to a stadium in a distant city. At noon, Brook

lyn is 367 miles away from the stadium and Nicole is 347 miles away from the stadium. Brooklyn is driving along the highway at a speed of 54 miles per hour and Nicole is driving at speed of 50 miles per hour. Let B represent Brooklyn's distance, in miles, away from the stadium t hours after noon. Let N represent Nicole's distance, in miles, away from the stadium t hours after noon. Write an equation for each situation, in terms of t, and determine the number hours after noon,t, when Brooklyn and Nicole are the same distance from the stadium.

1 answer:

3241004551 [841]3 years ago

3 0

Answer: B= -54t +367

N=-50t + 347

so in total would be 5 HOURS

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Pls help me for once....

ryzh [129]

The answers are C and also D

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3 years ago

1. an alloy contains zinc and copper in the ratio of 7:9 find weight of copper of it had 31.5 kgs of zinc.

m_a_m_a [10]

Answer:

Step-by-step explanation:

Question (1). An alloy contains zinc and copper in the ratio of 7 : 9.

If the weight of an alloy = x kgs

Then weight of copper = $\frac{9}{7+9}\times (x)$

= $\frac{9}{16}\times (x)$

And the weight of zinc = $\frac{7}{7+9}\times (x)$

= $\frac{7}{16}\times (x)$

If the weight of zinc = 31.5 kg

31.5 = $\frac{7}{16}\times (x)$

x = $\frac{16\times 31.5}{7}$

x = 72 kgs

Therefore, weight of copper = $\frac{9}{16}\times (72)$

= 40.5 kgs

2). i). 2 : 3 = $\frac{2}{3}$

4 : 5 = $\frac{4}{5}$

Now we will equalize the denominators of each fraction to compare the ratios.

$\frac{2}{3}\times \frac{5}{5}$ = $\frac{10}{15}$

$\frac{4}{5}\times \frac{3}{3}=\frac{12}{15}$

Since, $\frac{12}{15}>\frac{10}{15}$

Therefore, 4 : 5 > 2 : 3

ii). 11 : 19 = $\frac{11}{19}$

19 : 21 = $\frac{19}{21}$

By equalizing denominators of the given fractions,

$\frac{11}{19}\times \frac{21}{21}=\frac{231}{399}$

And $\frac{19}{21}\times \frac{19}{19}=\frac{361}{399}$

Since, $\frac{361}{399}>\frac{231}{399}$

Therefore, 19 : 21 > 11 : 19

iii). $\frac{1}{2}:\frac{1}{3}=\frac{1}{2}\times \frac{3}{1}$

$=\frac{3}{2}$

$\frac{1}{3}:\frac{1}{4}=\frac{1}{3}\times \frac{4}{1}$

= $\frac{4}{3}$

Now we equalize the denominators of the fractions,

$\frac{3}{2}\times \frac{3}{3}=\frac{9}{6}$

And $\frac{4}{3}\times \frac{2}{2}=\frac{8}{6}$

Since $\frac{9}{6}>\frac{8}{6}$

Therefore, $\frac{1}{2}:\frac{1}{3}>\frac{1}{3}:\frac{1}{4}$ will be the answer.

IV). $1\frac{1}{5}:1\frac{1}{3}=\frac{6}{5}:\frac{4}{3}$

$=\frac{6}{5}\times \frac{3}{4}$

$=\frac{18}{20}$

$=\frac{9}{10}$

Similarly, $\frac{2}{5}:\frac{3}{2}=\frac{2}{5}\times \frac{2}{3}$

$=\frac{4}{15}$

By equalizing the denominators,

$\frac{9}{10}\times \frac{30}{30}=\frac{270}{300}$

Similarly, $\frac{4}{15}\times \frac{20}{20}=\frac{80}{300}$

Since $\frac{270}{300}>\frac{80}{300}$

Therefore, $1\frac{1}{5}:1\frac{1}{3}>\frac{2}{5}:\frac{3}{2}$

V). If a : b = 6 : 5

$\frac{a}{b}=\frac{6}{5}$

$=\frac{6}{5}\times \frac{2}{2}$

$=\frac{12}{10}$

And b : c = 10 : 9

$\frac{b}{c}=\frac{10}{9}$

Since a : b = 12 : 10

And b : c = 10 : 9

Since b = 10 is common in both the ratios,

Therefore, combined form of the ratios will be,

a : b : c = 12 : 10 : 9

7 0

3 years ago

How would i do this problem and show my work x+9= -12

boyakko [2]

To solve this problem just do the following,
x+9=-12
-9 -9
x=-21
Check
-21+9=-12 = True

Hope this helps!=)

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3 years ago

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Suppose that det(a) = a b c d e f g h i = 2 and find the determinant of the given matrix. a b c −4d −4e −4f a + g b + h c + i

zhuklara [117]

I'll go out on a limb and suppose you're given the matrix

$\mathbf A=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}$

and you're asked to find the determinant of $\mathbf B$ , where

$\mathbf B=\begin{bmatrix}a&b&c\\-4d&-4e&-4f\\a+g&b+h&c+i\end{bmatrix}$

and given that $\det\mathbf A=2$ .

There are two properties of the determinant that come into play here:

(1) Whenever a single row/column is scaled by a constant $k$ , then the determinant of the matrix is scaled by that same constant;

(2) Adding/subtracting rows does not change the value of the determinant.

Taken together, we have that

$\det\mathbf B=-4\det\mathbf A=-8$

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3 years ago

What is the approximate volume of a cylinder with height of 2ft and radius of 6ft use 3.14 to approximate pi

DENIUS [597]

The answer to your problem is 226.08

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3 years ago