Let us take the second equation first
2a + 2b = 6
Dividing both sides by 2 we get
a + b = 3
a = 3 - b
Putting the value of a in the first equation we get
3a + 4b = 9
3(3 - b) + 4b = 9
9 - 3b + 4b = 9
b = 9 - 9
= 0
Now putting the value of b in the second equation we get
a + b = 3
a + 0 = 3
a = 3
So the value of the unknown variable a is 3 and the value of the unknown variable b is 0.
Answer:
The equation for area of a triangle is height multiplied by the base length, divided by 2. So, if the height is 2, and the base is 4, then you multiply the two together, making 8, and then divide it by 2, making the final area 4. So if the dimensions are doubled, then the equation would be 4 times 8 divided by 2. Which is 16. If we do this again with 16 times 8, we will get 128, which when divided by two is 64. This is 4 times 16, just as 16 is 4 times the original answer 4. This means that every time you double the dimensions you will get the original answer multiplied by 4.
Step-by-step explanation:

a. The gradient is


b. The gradient at point P(1, 2) is

c. The derivative of
at P in the direction of
is

It looks like

so that

Then


The answer to this question would be: <span>The new survey’s margin of error will be between 50% and 100% the size of the original survey’s margin of error.
A bigger sample will result in a narrower margin of error which is a good thing because your data will become more accurate. But twice size will not improve the margin into the half. It definitely became lower than 100% though
</span>
Answer:
£124
Step-by-step explanation:
first of all, you're meant to add all the figures given on the pie chart, which is 29+137+58+62+74 and that will give you a total of 360 degrees
so, you say 360 degrees is equal to the total amount of money he spent, which is £720, what about 62 degrees
therefore, you just do (62×720)÷360
that will give you £124