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3 years ago

How do I do this I'm so confused ??

Mathematics

1 answer:

exis [7]3 years ago

6 0

To find the slope you could use the rise/run method.

To do this you would select any 2 points that's on the graph.....Such as (2, 3) & (6, 4)

Then from the point (2,3), count up (rise) Then over (run) until you get to the other point (6, 4)

Which in this case is 1/4 (rise/run) which is the slope of the graph.

Hope this helps! :)

You might be interested in

One way to find a common denominator is by

Lena [83]

Answer:

Step-by-step explanation:

a). $1\frac{3}{5}=1+\frac{3}{5}$

$=1+\frac{3}{5}\times \frac{4}{4}$

$=1+\frac{12}{20}$

$=1\frac{12}{20}$

$1\frac{3}{4}=1+\frac{3}{4}$

$=1+\frac{3}{4}\times \frac{5}{5}$

$=1+\frac{15}{20}$

$=1\frac{15}{20}$

Therefore, common denominator of 20.

b). $2\frac{1}{2}=2+\frac{1}{2}$

$=2+\frac{1}{2}\times \frac{5}{5}$

$=2+\frac{5}{10}$

$=2\frac{5}{10}$

$\frac{4}{5}=\frac{4}{5}\times \frac{2}{2}$

$=\frac{8}{10}$

Common denominator of 10.

c). $\frac{3}{8}=\frac{3}{8}\times \frac{6}{6}$

= $\frac{18}{48}$

$\frac{1}{6}=\frac{1}{6}\times \frac{8}{8}$

= $\frac{8}{48}$

Common denominator of 48.

3 0

3 years ago

Find the value of y in this equation 16y=164

fiasKO [112]

16y = 164
y = 164 ÷ 16
y = 10.25

Hope This Helps You!

8 0

4 years ago

Read 2 more answers

Help Please! Easy Question I will mark brainest :)

Mashcka [7]

6. K=4
7. N=11
8. C=2
9. R=7
10. R=4
11. Y=1
12. R=7
13. N=7

3 0

3 years ago

The height, h, in feet of the tip of the hour hand of a wall clock varies from 9 feet to 10 feet. Which of the following equatio

bixtya [17]

The correct option is (b) h=0.5cos( $\pi$ /6t)+9.5.

The equations can be used to model the height as a function of time, t, in hours is h=0.5cos( $\pi$ /6t)+9.5.

<h3>Equation of cosine function:</h3>

The following is a presentation of the cosine function's generic form;

y = a + cos(bx - c) + d

amplitude = a

b = cycle speed

Calculation for the model height;

The height, h (feet) of the tip of the hour hand of a wall clock varies from 9 feet to 10 feet.

Obtain amplitude 'a' as

$\begin{aligned}a &=\frac{\text { Maximum value }-\text { Minimum value }}{2} \\a &=\frac{10-9}{2} \\a &=\frac{1}{2} \\a &=0.5\end{aligned}$

The time 'T' is calculated as-

$\begin{aligned}&\mathrm{T}=\frac{2 \pi}{\mathrm{b}} \\&12=\frac{2 \pi}{\mathrm{b}} \\&\mathrm{b}=\frac{2 \pi}{12} \\&\mathrm{~b}=\frac{\pi}{6}\end{aligned}$

Now, calculate 'd'

$\begin{aligned}&\mathrm{d}=\frac{\text { Maximum value }+\text { Minimum value }}{2} \\&\mathrm{~d}=\frac{10+9}{2} \\&\mathrm{~d}=\frac{19}{2} \\&\mathrm{~d}=9.5\end{aligned}$

Therefore, with the height as a function of time, t, expressed in hours, can be modeled by the following equations:

h=0.5cos( $\pi$ /6t)+9.5

To know more about the general equation of a cosine function, here

brainly.com/question/27587720

#SPJ4

The complete question is-

The height, h, in feet of the tip of the hour hand of a wall clock varies from 9 feet to 10 feet. Which of the following equations can be used to model the height as a function of time, t, in hours? Assume that the time at t = 0 is 12:00 a.m.

A. h=0.5cos( $\pi$ /12t)+9.5

B. h=0.5cos( $\pi$ /6t)+9.5

C. h=cos( $\pi$ /12t)+9

D. h=cos( $\pi$ /6t)+9

7 0

2 years ago

A cake is removed from a 310°F oven and placed on a cooling rack in a 72°F room. After 30 minutes the cake's temperature is 220°

Fynjy0 [20]

Answer:

The time is 135 min.

Step-by-step explanation:

For this situation we are going to use Newton's Law of Cooling.

Newton’s Law of Cooling states that the rate of temperature of the body is proportional to the difference between the temperature of the body and that of the surrounding medium and is given by

$T(t)=C+(T_0-C)e^{kt}$

where,

C = surrounding temp

$T(t)$ = temp at any given time

t = time

$T_0$ = initial temp of the heated object

k = constant

From the information given we know that:

Initial temp of the cake is 310 °F.
The surrounding temp is 72 °F.
After 30 minutes the cake's temperature is 220 °F.

We want to find the time, in minutes, since the cake's removal from the oven, at which its temperature will be 100°F.

To do this, first, we need to find the value of k.

Using the information given,

$220=72+(310-72)e^{k\cdot 30}\\\\72+238e^{k30}=220\\\\238e^{k30}=148\\\\e^{k30}=\frac{74}{119}\\\\\ln \left(e^{k\cdot \:30}\right)=\ln \left(\frac{74}{119}\right)\\\\k\cdot \:30=\ln \left(\frac{74}{119}\right)\\\\k=\frac{\ln \left(\frac{74}{119}\right)}{30}$

$T(t)=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}$

Next, we find the time at which the cake's temperature will be 100°F.

$100=72+(310-72)e^{(\frac{\ln \left(\frac{74}{119}\right)}{30}\cdot t)}\\72+238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=100\\238e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=28\\e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}=\frac{2}{17}\\\ln \left(e^{\frac{\ln \left(\frac{74}{119}\right)}{30}t}\right)=\ln \left(\frac{2}{17}\right)\\\frac{\ln \left(\frac{74}{119}\right)}{30}t=\ln \left(\frac{2}{17}\right)\\t=\frac{30\ln \left(\frac{2}{17}\right)}{\ln \left(\frac{74}{119}\right)}\approx 135.1$

4 0

3 years ago