Are the given angles complementary, supplementary, or neither? ∠VYU and ∠UYZ

A. complementary

B. supplementary

C. neither

Answers

Answer 1

Answer: B. Complementary means totaling 90 degrees, and supplementary means totaling 180 degrees.

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a bank contains 27 coins consisting of dimes and quarters. the coins have a total value of $4.95. Find the number of dimes and quarters in the bank.
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Joshua delivered 30 hives to the local fruit farm if the farmer has paid to use 5 percent of the total number of Joshua's hives how many hives does Joshua have in all

Answers

so if Joshua brought 30 hives and that is 5% of all of his hives
then 30=5/100
we know that 5/100 simplifies to 1/20 so if
1/20=30 then we multiply both sides by 20 to get
20/20 (1/1 or 100%) = 600
so Joshua has 600 hives in all

The sum of two consecutive even interfere is -10. what is the large integer?

Answers

-4 —> -4 + -6 = -10, so -4 is greater as it is less negative

What is a dollar in fraction

Answers

1 because if you think of it like money 1 doolar is 1 whole. Same goes with fractions

Another way to look at this is to think: 6 dimes makes 60 cents while a whole dollar is 100 cents. So the fraction is 60%2F100 which will reduce to 6%2F10 and then finally reduce to 3%2f5

How do you do One-Step Equations?

Answers

I agree with the comments. If you give an example, I will walk you through the steps to solve it.

It depends on the kind of equation. It could be subtraction, addition, division, our multiplication. If you give an equation, I may be able to help you.

Evaluate 1/6 + 2/3. a.1/5 b.3/18 c.5/6 d.1 1/6

Answers

Answer:

5/6

Step-by-step explanation:

1/6 + 2/3

We need to get a common denominator of 6

1/6 + 2/3 *2/2

1/6 + 4/6

5/6

100 POINTS HELP PLEASEEEE !!!HURRY!!!

Answers

Question # 1 Solution

Answer:

$(h-f)(-8)=-229$

Step-by-step Solution:

Given

$h(r)=-4r^(2)+4$

$f(r)=2r-7$

$(h-f)(r)=h(r)-f(r)$

$(h-f)(r)=-4r^(2)+4-(2r-7)$

$(h-f)(r)=-4r^(2)+4-2r+7)$

We have to find $(h-f)(-8)$

So,

$(h-f)(-8)=-4(-8)^(2)+4-2(-8)+7)$

$(h-f)(-8)=-4(64)+4+16+7$

$(h-f)(-8)=-256+4+16+7$

$(h-f)(-8)=-229$

∴ $(h-f)(-8)=-229$

Question # 2 Solution

Answer:

∴ $(f.p)(9)=38979$

Step-by-step Solution:

Given

$f(s)=7s-2$

$p(s)=s^(3)-10s$

$(f.p)(s)=f(s).p(s)$

$(f.p)(s)=(7s-2)(s^(3)-10s)$

$(f.p)(s)=7s^4-2s^3-70s^2+20s$

We have to find $(f.p)(9)$

So,

$(f.p)(9)=7\left(9\right)^4-2\left(9\right)^3-70\left(9\right)^2+20\left(9\right)$

$(f.p)(9)=7\cdot \:9^4-2\cdot \:9^3-70\cdot \:9^2+20\cdot \:9$

$(f.p)(9)=9^4\cdot \:7-9^3\cdot \:2-9^2\cdot \:70+180$

$(f.p)(9)=45927-1458-5670+180$

$(f.p)(9)=38979$

∴ $(f.p)(9)=38979$

Question # 3 Solution

Answer:

$((f)/(p))(r)=(11r+8)/(r\left(r^2+6\right))$

Step-by-step Solution:

Given

$p(r)=r^(3)+6r$

$f(r)=11r+8$

We have to find $((f)/(p))(r)$

Using the formula

$((f)/(p))(r)=(f(r))/(p(r))$

$p(r)=r^(3)+6r$

$f(r)=11r+8$

$((f)/(p))(r)=(11r+8)/(r^(3)+6r)$

$((f)/(p))(r)=(11r+8)/(r\left(r^2+6\right))$

∴ $((f)/(p))(r)=(11r+8)/(r\left(r^2+6\right))$

Question # 4 Solution

Answer:

$(h-p)(k)=5k^2-3-k^3-8k$

Step-by-step Solution:

Given

$h(k)=5k^(2)-3$

$p(k)=k^(3)+8k$

We have to find $(h-p)(k)$

Using the formula

$(h-p)(k)=h(k)-p(k)$

$h(k)=5k^(2)-3$

$p(k)=k^(3)+8k$

$(h-p)(k)=5k^(2)-3-(k^(3)+8k)$

$(h-p)(k)=5k^2-3-k^3-8k$

∴ $(h-p)(k)=5k^2-3-k^3-8k$

Question # 5 Solution

Answer:

$((p)/(g))(11)=(1287)/(155)$

Step-by-step Solution:

Given

$p(b)=b^(3)-4b$

$g(b)=b^(2)+4b-10$

We have to find $((p)/(g))(11)$

$((p)/(g))(b)=(p(b))/(g(b))$

$((p)/(g))(b)=(b^(3)-4b)/(b^(2)+4b-10)$

$((p)/(g))(11)=(11^(3)-4(11))/(11^(2)+4(11)-10)$

$((p)/(g))(11)=(1287)/(155)$

∴ $((p)/(g))(11)=(1287)/(155)$

Question # 6 Solution

Answer:

$(f+g)(x)=x^2+20x-18$

Step-by-step Solution:

Given

$g(x)=x^(2)+11x-7$

$f(x)=9x-11$

We have to find $(f+g)(x)$

$(f+g)(x)=f(x)+g(x)$

$(f+g)(x)=9x-11+(x^(2)+11x-7)$

$(f+g)(x)=9x-11+x^(2)+11x-7$

$(f+g)(x)=x^2+20x-18$

∴ $(f+g)(x)=x^2+20x-18$

Question # 7 Solution

Answer:

$h(10)+g(10)=-983$

Step-by-step Solution:

Given

$h(w)=-11w^(2)-7$

$g(w)=w^(2)+3w-6$

We have to find $h(10)+g(10)$

So,

$h(10)=-11(10)^(2)-7$

$h(10)=-1107$ .....[1]

and

$g(10)=10^(2)+3(10)-6$

$g(10)=124$ .....[2]

Adding Equation [1] and Equation [2]

$h(10)+g(10)=-1107+124$

$h(10)+g(10)=-983$

∴ $h(10)+g(10)=-983$

Question # 8 Solution

Answer:

$(f+g)(-3)=-6$

Step-by-step Solution:

Given

$g(b)=b^(2)+9b+10$

$f(b)=3b+11$

We have to find $(f+g)(-3)$

$(f+g)(b)=f(b)+g(b)$

$(f+g)(b)=3b+11+b^(2)+9b+10$

$(f+g)(-3)=3(-3)+11+(-3)^(2)+9(-3)+10$

$(f+g)(-3)=-6$

∴ $(f+g)(-3)=-6$

Question # 9 Solution

Answer:

$(f.h)(k)=33k^3+22k-12k^2-8$

Step-by-step Solution:

Given

$f(k)=-11k+4$

$h(k)=-3k^(2)-2$

We have to find $(f.h)(k)$

$(f.h)(k)=f(k).h(k)$

So,

$(f.h)(k)=(-11k+4).(-3k^(2)-2)$

$\mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$

∵ FOIL means (First, Outer, Inner, Last)

$(f.h)(k)=\left(-11k\right)\left(-3k^2\right)+\left(-11k\right)\left(-2\right)+4\left(-3k^2\right)+4\left(-2\right)$

$(f.h)(k)=33k^3+22k-12k^2-8$

∴ $(f.h)(k)=33k^3+22k-12k^2-8$

Question # 10 Solution

Answer:

$p(-8)-f(-8)=-581$

Step-by-step Solution:

Given

$p(s)=s^(3)+6s$

$f(s)=-2s+5$

We have to find $p(-8)-f(-8)$

So,

$p(-8)=(-8)^(3)+6(-8)$

$p(-8)=-560.....[1]$

and

$f(-8)=-2(-8)+5$

$f(-8)=21.....[2]$

Subtracting Equation [2] from Equation [1]

$p(-8)-f(-8)=-560-21$

$p(-8)-f(-8)=-581$

∴ $p(-8)-f(-8)=-581$

Keywords: function operation

Learn more about function operations from brainly.com/question/3725682

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Answer:

-2229

Step-by-step explanation:

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Are the given angles complementary, supplementary, or neither? ∠VYU and ∠UYZ A. complementary B. supplementary C. neither

Answers

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Are the given angles complementary, supplementary, or neither? ∠VYU and ∠UYZ

A. complementary

B. supplementary

C. neither