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2 years ago

1. How many high fives will alan get in 30 seconds?

Mathematics

2 answers:

Mamont248 [21]2 years ago

4 0

I guess 24 in 30sec on completing one round each

defon2 years ago

3 0

Answer:

1) $Six$

Step-by-step explanation:

1) $30\ seconds\div5\ seconds=6\ seconds$ $\left\{every\ 5\ seconds\ one\ time\right\}$

I am going to work on the other several tonight/tomorrow and will post when I get them done.

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Can someone please help me with this? Thanks

Nata [24]

The answer is 2x+9+4 over x-2

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2 years ago

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soldi70 [24.7K]

Answer:

solution given:

CE=2×2=4[base side is double to the line of mid point]

AB=15/2=7.5[b is a mid point]

m<AEC=75°[ corresponding angle]

given:

BD=x+1

CE=x+10

we have

2BD=CE

2(x+1)=x+10

2x+2=x+10

2x-x=10-2

x=8

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2 years ago

5b +20y<br> A. 5(b+4y)<br> B. 5(4y+b)<br> C. 2(b+10y)

enot [183]

Answer:

a once again it is easy the answer is A

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Solve each equation for the indicated variable. u = -3a - 3 , for a

MAXImum [283]

Answer:

$a = \frac{u + 3}{-3}$

Step-by-step explanation:

Isolate the variable, a. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS.

PEMDAS is the order of operation and stands for:

Parenthesis

Exponents (& Roots)

Multiplication

Division

Addition

Subtraction

First, add 3 to both sides of the equation:

u = -3a - 3

u (+3) = -3a - 3 (+3)

u + 3 = -3a

Next, divide -3 from both sides of the equation:

(u + 3)/-3 = (-3a)/-3

a = (u + 3)/-3

$a = \frac{u + 3}{-3}$ is your answer.

8 0

2 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

2 years ago