Answer:
7.8
Step-by-step explanation:
Answer:26
Step-by-step explanation:
The reason i got 26:
-we substitute the letters(variables) and have 52=(k)2
but what is k? k is y divided by x (just like y is k times x)
Martin's answer is wrong. The afternoon temperature in Westfield city is -10°F.
Given,
The temperature in Westfield city in morning = -2°F
The drop down temperature by afternoon = 8°F
We have to find the temperature in afternoon:
Temperature in afternoon = Temperature in morning - Drop down temperature
= -2 - 8 = -10
The afternoon temperature in Westfield city is -10°F
Martin subtracted like:
-2 - (-8) = -2 + 8 = 6
This is wrong.
That is,
We can conclude like:
Martin's answer is wrong. The temperature in Westfield city in afternoon is -10°F
Learn more about temperature here:
brainly.com/question/20855720
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Step-by-step explanation:
All you gotta do is multiply all the test amounts by 5 1/2.
If you have a calculator handy, I would change everything to decimals.
So for chopped onions, you would need
1.5 lbs x 5.5 = 8.25 = 8 1/4 lbs of chopped onion.
Chopped green pepper
.5 lbs x 5.5 = 2.75 = 2 3/4 lbs chopped onion.
Carrots
.5 lbs c 5.5 = 3.75 = 2 3/4 lbs carrots
Chopped Celery
4 stalks x 5.5 = 22 stalks of chopped celery.
And so on. I think you can do it from there.
Answer:
(-1,5) and
are points on the graph
Step-by-step explanation:
Given

Required
Determine which point in on the graph
To get which of point A to D is on the graph, we have to plug in their values in the given expression using the format; (x,g(x))
A. (-1,5)
x = -1
Substitute -1 for x in 

Convert to index form

Change / to *


This satisfies (-1,5)
<em>Hence, (-1,5) is on the graph</em>
<em></em>
B. (1,0)
x = 1
Substitute 1 for x



<em>(1,0) is not on the graph because g(x) is not equal to 0</em>
C. 
x = 3
Substitute 3 for x


Apply law of indices


This satisfies 
<em>Hence, </em>
<em> is on the graph</em>
<em></em>
D. 
x = -2
Substitute -2 for x


Convert to index form



Change / to *


This does not satisfy 
<em>Hence, </em>
<em> is not on the graph</em>