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8_murik_8 [283]

3 years ago

5

Line segment AC where point B is located between A and C solve if AB=39 and AC=46 find BC

1 answer:

jeka943 years ago

6 0

Answer:

BC = 7

Step-by-step explanation:

AB + BC = AC

39 + BC = 46

BC = 46 - 39

BC = 7

You might be interested in

Suppose that the functions u and w are defined as follows.

jasenka [17]

Answer(there are assumptions for this answer that you need to confirm and look at):

Assumptions: $u(x)=x^2+3$ and $w(x)=\sqrt{x+2}$

Answer if the operation is multiplication:

If you meant a closed dot which is the symbol for multiplication.

$(u \cdot w)(2)=14$

$(w \cdot u)(2)=14$

Answer if the operation is composition:

If you meant an open dot which is the symbol for composition.

$(u \circ w)(2)=7$

$(w \circ u)(2)=3$

Note: I don't know if you actually meant $w(x)=\sqrt{x+2}$ or if $w(x)=\sqrt{x}+2$ . Please let me know one way or the other.

Step-by-step explanation:

If we assume the functions are:

$u(x)=x^2+3$

$w(x)=\sqrt{x+2}$

$u \cdot w=w \cdot u$ since multiplication is commutative.

$u(2)=2^2+3$

$u(2)=4+3$

$u(2)=7$

$w(2)=\sqrt{2+2}$

$w(2)=\sqrt{4}$

$w(2)=2$

We are asked to find $(u \cdot w)(2)$ and $(w \cdot u)(2)$ .

The order doesn't matter in multiplication.

$(u \cdot w)(2)$

$u(2) \cdot w(2)$

$7 \cdot 2$

$14$

$(w \cdot u)(2)$

$w(2) \cdot u(2)$

$2 \cdot 7$

$14$

Now you might have meant composition which symbolized with an open circle, not a closed one.

$(u \circ w)(2)$

$u(w(2))$

$u(2)$ since $w(2)=2$

$2^2+3$

$4+3$

$7$

$(w \circ u)(2)$

$w(u(2))$

$w(7)$ since $u(2)=7$

$\sqrt{7+2}$

$\sqrt{9}$

$3$

6 0

4 years ago

Polygon ABCD was transformed in polygon A'B'C'D. Identify the sequence of transformations performed. Select all that applied

stich3 [128]

Answer:

B. Polygon ABCD was reflected over the x-axis and then reflected over the y-axis.

6 0

3 years ago

Answer number 36 plz ✨

Nikitich [7]

it is d. d is the answer

8 0

3 years ago

18x2 + 10x = 0<br> please help no spam!

9966 [12]

Answer:

x= -5/9

That is the answer, try it.

4 0

3 years ago

A glitter glue recipe takes a mixture of 1 5 cup of glitter to 2 5 cup of glue. For A, how many cups of glitter are used for 1 c

tatiyna

0.6 cup of gliter is used for 1 cup of glue and 1.6 cup of gliter glue is made with 1 cup of glue.

Part A

Given that,

15 cup of gliter is for 25 cup of glue ,

Therefore, "X" cup of gliter is for 1 cup of glue

X = (1*15)/25

X= 0.6 cup of gliter

Part B,

Given that,

From part A 0.6 cup of gliter is for 1 cup of glue

Thus, 1 cup of glue and 0.6 cup of gliter when mixed will form "Y" cup of gliter glue

Now, Y= 0.6 cup of gliter + 1 cup of glue

Y= 1.6 cup of gliter glue

which approximates to 2 cup of gliter glue.

For more information on cross multiplication kindly visit to

brainly.com/question/11203238

3 0

1 year ago