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2 years ago

8

Solve the equation 13x+16y=11 for y

1 answer:

Nana76 [90]2 years ago

6 0

x − 3 = 6 − 2 x = − 8 √ x = 3

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One leg of a right triangle is 28 inches longer than the other leg, and the hypotenuse is 52 inches. Find the lengths of the leg

katovenus [111]

Step-by-step explanation:

Let x represent the missing sides measurement.

x+x+28=52

combine like terms

2x + 28 = 52

subtract 28 from both sides

2x + 28 = 52

- 28 - 28

2x = 24

divide both sides by 2

x = 12

plug it back in

one leg is 12 inches

the other leg is 40 inches

7 0

2 years ago

If a line has a slope of -4 and a u intercept of 1, which of the following is it's graph?

3241004551 [841]

Well. I can't see the picture. But it would be decreasing at a rate of 4 so pick a point on the any line move over to the right one, then down four. And for the y intercept the line would intersect the y axis at positive one. Sorry if that's confusing

3 0

3 years ago

Convert 31 minutes to seconds

In-s [12.5K]

Answer:

Step-by-step explanation: 1,860 Seconds Is Equal To 31 Minutes Hope It Helped

6 0

3 years ago

Evaluate the function at each specified value of the independent variable and simplify. (if an answer is undefined, enter undefi

tekilochka [14]

For this case we have the following function:

$h (t) = t ^ 2 - 8t$

What we should do is evaluate the function for the different values of the independent variable.

We have then:

For x = 8:

$h (8) = (8) ^ 2 - 8 (8)$

$h (8) = 64 - 64$

$h (8) = 0$

For x = 1.8:

$h (1.8) = (1.8) ^ 2 - 8 (1.8)$

$h (1.8) = 3.24 - 14.4$

$h (1.8) = -11.16$

For x = x + 8:

$h (x + 8) = (x + 8) ^ 2 - 8 (x + 8)$

$h (x + 8) = x ^ 2 + 16x + 64-8x-64$

$h (x + 8) = x ^ 2 + 18x$

Answer:

The results of evaluating the function are:

$h (8) = 0$

$h (1.8) = -11.16$

$h (x + 8) = x ^ 2 + 18x$

5 0

2 years ago

1. Let a; b; c; d; n belong to Z with n > 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition

lukranit [14]

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) $a \equi b (mod n) \rightarrow a-b=kn$ for integer $k$ .

1) $c \equi d (mod n) \rightarrow c-d=mn$ for integer $m$ .

a)

Proof:

We want to show $a+c \equiv b+d (mod n)$ .

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(a+c)-(b+d)=rn. If we do that we would have shown that $a+c \equiv b+d (mod n)$ .

kn+mn = (a-b)+(c-d)

(k+m)n = a-b+ c-d

(k+m)n = (a+c)+(-b-d)

(k+m)n = (a+c)-(b+d)

k+m is is just an integer

So we found integer r such that (a+c)-(b+d)=rn.

Therefore, $a+c \equiv b+d (mod n)$ .

//

b) Proof:

We want to show $ac \equiv bd (mod n)$ .

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(ac)-(bd)=tn. If we do that we would have shown that $ac \equiv bd (mod n)$ .

If a-b=kn, then a=b+kn.

If c-d=mn, then c=d+mn.

ac-bd = (b+kn)(d+mn)-bd

= bd+bmn+dkn+kmn^2-bd

= bmn+dkn+kmn^2

= n(bm+dk+kmn)

So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.

Therefore, $ac \equiv bd (mod n)$ .

//

3 0

3 years ago