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sleet_krkn [62]
3 years ago
9

Question number 4, solve the formula for the specified variable

Mathematics
2 answers:
Wittaler [7]3 years ago
6 0

Answer:

the answer is c = 5/9(F-32)

svlad2 [7]3 years ago
4 0
I’m not sure maybe it’s A..?
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ABCD is a quadrilateral. P is a point. P is closer to C than to D. P is also closer tothe line AB than the line AD. Find and sha
babymother [125]

Answer:

APBCD

Step-by-step explanation:

Because this make sance

4 0
3 years ago
Pia printed 2 maps of a walking trail. The length of the trail on the first map is 8 cm. The length of the trail on the second m
Marianna [84]
The scale factor from the first map to the second map can be determined by:

scale factor = (length of trail in first map) / (length of trail in second map)

8 cm / 6 cm = 1.33

The landmark on the first map is a triangle with side lengths of 3mm, 4mm, and 5mm. 

3mm = 3 * 1.33 mm = 3.99mm
4mm = 4 * 1.33 mm = 5.32mm
5mm = 5 * 1.33 mm = 6.65mm

5 0
2 years ago
What is the area of this irregular figure?
VikaD [51]

Answer:

hey man if you figure this out comment back bc i need to know also. sorry and thanks!

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
MATH HELP PLS!!! MARKING BRAINLIESTT!!!!!!!!!!!!!!!!!
kow [346]

The function would be y = 0.45x, x being the number of seconds, y being the items he can ring up.

For example (this is an example only to show how this equation works), if we were looking at how many items the cashier can ring in 5 seconds, we would make the equation y = 0.45(5), and with multiplication, we can find that they would ring 2.25 items ( y = 2.25 )

If I misunderstood this question or got something wrong, please leave a comment and I can help further.

8 0
2 years ago
If f(1) = 0, what are all the roots of the function f(x)=x^3+3x^2-x-3? Use the Remainder Theorem.
Sophie [7]
There's no if about it, 

f(x)=x^3+3x^2-x-3


has a zero f(1)=0 so x-1 is a factor.   That's the special case of the Remainder Theorem; since f(1)=0 we'll get a remainder of zero when we divide f(x) by x-1.

At this point we can just divide or we can try more little numbers in the function.  It doesn't take too long to discover f(-1)=0 too, so  x+1 is a factor too by the remainder theorem.  I can find the third zero as well; but let's say that's out of range for most folks.

So far we have 

x^3+3x^2-x-3 = (x-1)(x+1)(x-r)

where r is the zero we haven't guessed yet.  Again we could divide f(x) by (x-1)(x+1)=x^2-1 but just looking at the constant term we must have

-3 = -1 (1)(-r) = r

so

x^3+3x^2-x-3 = (x-1)(x+1)(x+3)

We check f(-3)=(-3)^3+3(-3)^2 -(-3)-3 = 0 \quad\checkmark

We usually talk about the zeros of a function and the roots of an equation; here we have a function f(x) whose zeros are

x=1, x=-1, x=-3

8 0
2 years ago
Read 2 more answers
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