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zloy xaker [14]
3 years ago
9

Plss help Given the equation :

Mathematics
2 answers:
Leya [2.2K]3 years ago
8 0
X=0 i really hope this helps
Nata [24]3 years ago
4 0

Answer:

   x=0

Step-by-step explanation:

5+x-12=2x-7

-12+5=-7

x-7=2x-7

-x    -x

x-7=-7

+7    +7

x=0

for b substitute the answer in to the equation

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in a sale there is twenty five per cent off all prices a bed cost £33 in the sale how much was it before the sale ?
goldfiish [28.3K]
The original price was 100% and the sale was -25%
100% - 25% = 75%
This means 75% = £33
75% = 3/4
If we divide the sale amount by 3 and them multiply by 4, we will get the original amount.
£33 / 3 = £11
£11 x 4 = £44
5 0
3 years ago
Two photos were taken of a car.
LenKa [72]

Answer:

14.4 m / s

Step-by-step explanation:

=distance between the car

=0.8 × 9

=7.2m

=time taken

=0.5s

=speed

=distance/time

=7.2m/0.5s

=14.4 m / s

4 0
2 years ago
Hello I need help with this someone please get back too me (I’ll give you 50 points) the problem says Estimate. Then Measure eac
andreyandreev [35.5K]

Answer:

The crayon is about 3 times longer than the paper clip.

The nearest inch: is 3

The nearest 1/2 inch, 3 1/2

The nearest 1/4 inch, 3 1/4

4 0
3 years ago
18 55 -101 196 55 18 22 X-22 57 88
choli [55]

Answer:

maybe a that's my answer

8 0
3 years ago
I need help please. Thanks!
Karolina [17]

Answer:

A

Step-by-step explanation:

We are given the function f and its derivative, given by:

f^\prime(x)=x^2-a^2=(x-a)(x+a)

Remember that f(x) is decreasing when f'(x) < 0.

And f(x) is increasing when f'(x) > 0.

Firstly, determining our zeros for f'(x), we see that:

0=(x-a)(x+a)\Rightarrow x=a, -a

Since a is a (non-zero) positive constant, -a is negative.

We can create the following number line:

<-----(-a)-----0-----(a)----->

Next, we will test values to the left of -a by using (-a - 1). So:

f^\prime(-a-1)=(-a-1-a)(-a-1+a)=(-2a-1)(-1)=2a+1

Since a is a positive constant, (2a + 1) will be positive as well.

So, since f'(x) > 0 for x < -a, f(x) increases for all x < -a.

To test values between -a and a, we can use 0. Hence:

f^\prime(0)=(0-a)(0+a)=-a^2

This will always be negative.

So, since f'(x) < 0 for -a < x < a, f(x) decreases for all -a < x < a.

Lasting, we can test all values greater than a by using (a + 1). So:

f^\prime(a+1)=(a+1-a)(a+1+a)=(1)(2a+1)=2a+1

Again, since a > 0, (2a + 1) will always be positive.

So, since f'(x) > 0 for x > a, f(x) increases for all x > a.

The answer choices ask for the domain for which f(x) is decreasing.

f(x) is decreasing for -a < x < a since f'(x) < 0 for -a < x < a.

So, the correct answer is A.

3 0
3 years ago
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