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Anastaziya [24]
2 years ago
14

2 (10- 22x0) = (2x + 2)

Mathematics
1 answer:
Lorico [155]2 years ago
3 0

Answer:

x=9

Step-by-step explanation:

2(10-22×0)=(2x+2)

Applying BODMAS

2(10-0)=(2x+2)

2(10)=2x+2

2x+2=20

2x=20-2

2x=18

x=9

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The length of a rectangular prism is 12 inches, and its width is 8 inches. The height of the prism is 16 inches. What is the sur
finlep [7]

Answer:

The answer to your question is letter B

Step-by-step explanation:

Data

length = 12 in

width = 8 in

height = 16 in

surface area = ?

Formula

Surface area = 2( length x width) + 2(length x height) + 2((width x height)

Substitution

Surface area = 2(12 x 8) + 2(12 x 16) + 2(8 x 16)

Simplification

Surface area = 256 + 384 + 102

Result

Surface area = 832 in²

7 0
3 years ago
Rewrite the equation in logarithmic form: 4^3 = 64
storchak [24]
4^3 = 64

log4 64 = 3

The exponent always comes to the other side.
8 0
3 years ago
Deka measured the mass of 100 sheets of paper as 1,500 grams. How many kilometers would 1 sheet of paper weigh?
Ray Of Light [21]
100=1500g so 10=150g so 1=15g
3 0
3 years ago
Read 2 more answers
From a boat on the lake, the angle of elevation to the top of a cliff is 26°1'. If the base of the cliff is 205 feet from the bo
e-lub [12.9K]
First off hmm <span>26°1' is just 26 degrees and 1 minute

there are 60 minutes in 1 degree, so 1minute is just 1/60 degrees
so </span>26°1' is 26° + (1/60)°  or 26.1667 rounded up, since 1/60 is a recurring unit

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8 0
2 years ago
The picture shows a triangular island:
sergiy2304 [10]

Answer:

The expressions that show the value of q are

1) q=\sqrt{r^{2}+s^{2}}

2) q=\frac{s}{cos(55\°)}

3) q=\frac{r}{sin(55\°)}

4) q=\frac{s}{sin(35\°)}

5) q=\frac{r}{cos(35\°)}

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

case A)

In the right triangle of the figure

Applying the Pythagoras Theorem

q^{2}=r^{2}+s^{2}

q=\sqrt{r^{2}+s^{2}}

case B)

In the right triangle of the figure

cos(55\°)=\frac{s}{q}

solve for q

q=\frac{s}{cos(55\°)}

case C)

In the right triangle of the figure

sin(55\°)=\frac{r}{q}

solve for q

q=\frac{r}{sin(55\°)}

case D)

In a right triangle

if A+B=90\°

then

cos(A)=sin(B)

therefore

q=\frac{s}{cos(55\°)}------> q=\frac{s}{sin(35\°)}

q=\frac{r}{sin(55\°)} ------>  q=\frac{r}{cos(35\°)}

4 0
3 years ago
Read 2 more answers
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