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3 years ago

Hellppppp pleaseeeee

Mathematics

1 answer:

Eduardwww [97]3 years ago

6 0

Answer:

The answer would be C,

Yes the graph passes the vertical line test

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Which equation represents the solution of the linear equation - 4x + 5 = 9?

Ludmilka [50]

The answer is C. C=-1

-4(-1)= 4 a negative times a negative is positive
4+5 is 9

5 0

2 years ago

Read 2 more answers

Daniela and Jack are hiking a steady incline. They use their GPS device to determine their elevation every 15 minutes. At 15 min

kow [346]

Answer:

$h=40t+9700$

where the elevation, h, is in feet and time, t, is in minutes.

Step-by-step explanation:

At the time, $t_1=15$ min, the elevation, $h_1= 10,300$ feet and

at the time $t_2= 30$ min, the elevation, $h_2=10,900$ feet.

As they are hiking a steady incline, so the change in the elevation with respect to time will be constant.

So, there will be a linear relationship between the elevation and the time.

Let $h$ be the elevation at any time instant $t$ , so the linear relation among these quantities is

$h=mt+C_0\;\cdots(i)$

where $m$ is the rate of change of elevation with respect to time and $C_0$ is constant.

The change in the elevation, $\Delta h=h_2-h_1= 10,900-10,300=600$ feet.

and the change in time, $\Delta t=t_2-1_1=30-15=15$ min.

So, change in the elevation in unit time,

$m=\frac{\Delat h}{\Delta t}=\frac{600}{15}=40$ feet/min.

Now, from equation (i)

$h=40t+C_0$

As the elevation, $h=h_1$ at time $t=t_1$ , so

$10,300=40\times15+C_0$

$\Rightarrow C_0=9700$

Hence, the required equation is

$h=40t+9700$

where the elevation, $h$ , is in feet and time, $t$ , is in minutes.

3 0

3 years ago

I dont know how to do this so hElp

STALIN [3.7K]

Answer:

3.5× c20

#######################

7 0

2 years ago

Read 2 more answers

If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

$b) log_b 324$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

$c) log_b\frac{8}{9}$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

$d) \frac{log_b27}{log_b16}$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$

Thus the given are evaluated in terms of x and y

3 0

3 years ago

Joe's final exam has true/false questions, worth 2 points each, and multiple choice questions, worth 5 points each. Let x be the

mixer [17]

This is the inequality

2x + 5y ≥ 90

7 0

3 years ago