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gavmur [86]
3 years ago
14

If a+ar =b+r the value of an item of b and r can be expressed as

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
5 0
a+ar=b+r\\
a(1+r)=b+r\\
a=\dfrac{b+r}{1+r}
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У = 3х – 1<br>у = 3y= -13<br>​
Elina [12.6K]

Answer:

y=1/3x-1/3

Step-by-step explanation:

6 0
3 years ago
HELP PLEASE EQUATION OF PARABOLA
Masteriza [31]

It has a minimum value at x = 3 and f(x) = 4

Vertex form is

f(x) = a(x - 3)^2 + 4 where a is some constant to be found

From the graph when x = 5 f(x) = 15, so

15 = a * 2^2 + 4

a = 15-4/4 = 11/4

so our equation is f(x) = 11/4(x - 3)^2 + 4

6 0
3 years ago
Use Pythagorean theorem!!! Will give b !!
aliina [53]

Answer:

The last one

Step-by-step explanation:

Hey There!

The Pythagorean Theorem states that

the two legs squared added together equal the hypotenuse square

we are given the two legs so to find the missing side we square them, then add them together

1^2=1\\3^2=9\\9+1=10

the missing side length equals

\sqrt{10}

hope this helps!

4 0
3 years ago
Which statement comparing ​|−12|​ and ​|8|​ is true?Which statement comparing ​|−12|​ and ​|8|​ is true?
NARA [144]

Answer:

The second one

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Let u = (1,2), v = (−3,4), and w = (5,0)
sergij07 [2.7K]

Answer:

w-2u-v

Step-by-step explanation:

Given are three vectors u, v and w.

In R^2 we treat first element as x coordinate and 2nd element as y coordinate.

Thus we mark (1,2) in the I quadrant, (-3,4) in II quadrant and (5,0) on positive x axis 5 units form the origin.

b) w=au+bv

We have to find the values of a and b

(5,0) = a(1,2)+b(-3,4)]

Equate the corresponding terms

5=a-3b\\0=2a+4b

Divide II equation by 2 to get

0=a+2b

Eliminate a

-5 = 5b: b=-1

a=-2

Hence

w = 2u-v

7 0
3 years ago
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