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ivanzaharov [21]

3 years ago

9

What is four and ninety two hundredths

2 answers:

Kryger [21]3 years ago

5 0

4.92.

4 is a whole number, so it goes in the ones place
The ths means it is going to be a decimal, so 92 goes two right of the decimal point. Hope this helped

grandymaker [24]3 years ago

3 0

I believe it is 4.92 .....

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2t-7-5t=11 what is the answer

Verizon [17]

<span>2t-7-5t=11

Start by combing like terms:

2-5t=-3t

Now we have:

-3t-7=11

Add 7 to both sides

-3t=18

To isolate t we divide both sides by -3

Final answer:
t=-6 </span>

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3 years ago

Find the slope of the line passing through each of the following pairs of points and draw the graph of the line.

tiny-mole [99]

Answer:

The slope is 1/2

Step-by-step explanation:

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Another Quick Question, True or false: The Intersection of two rays creates two pairs of vertical angles and four pairs of suppl

andrey2020 [161]

Answer:

True

Step-by-step explanation:

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3 years ago

What is the solution of

kobusy [5.1K]

Answer:

Third option: x=0 and x=16

Step-by-step explanation:

$\sqrt{2x+4}-\sqrt{x}=2$

Isolating √(2x+4): Addind √x both sides of the equation:

$\sqrt{2x+4}-\sqrt{x}+\sqrt{x}=2+\sqrt{x}\\ \sqrt{2x+4}=2+\sqrt{x}$

Squaring both sides of the equation:

$(\sqrt{2x+4})^{2}=(2+\sqrt{x})^{2}$

Simplifying on the left side, and applying on the right side the formula:

$(a+b)^{2}=a^{2}+2ab+b^{2}; a=2, b=\sqrt{x}$

$2x+4=(2)^{2}+2(2)(\sqrt{x})+(\sqrt{x})^{2}\\ 2x+4=4+4\sqrt{x}+x$

Isolating the term with √x on the right side of the equation: Subtracting 4 and x from both sides of the equation:

$2x+4-4-x=4+4\sqrt{x}+x-4-x\\ x=4\sqrt{x}$

Squaring both sides of the equation:

$(x)^{2}=(4\sqrt{x})^{2}\\ x^{2}=(4)^{2}(\sqrt{x})^{2}\\ x^{2}=16 x$

This is a quadratic equation. Equaling to zero: Subtract 16x from both sides of the equation:

$x^{2}-16x=16x-16x\\ x^{2}-16x=0$

Factoring: Common factor x:

x (x-16)=0

Two solutions:

1) x=0

2) x-16=0

Solving for x: Adding 16 both sides of the equation:

x-16+16=0+16

x=16

Let's prove the solutions in the orignal equation:

1) x=0:

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(0)+4}-\sqrt{0}=2\\ \sqrt{0+4}-0=2\\ \sqrt{4}=2\\ 2=2$

x=0 is a solution

2) x=16

$\sqrt{2x+4}-\sqrt{x}=2\\ \sqrt{2(16)+4}-\sqrt{16}=2\\ \sqrt{32+4}-4=2\\ \sqrt{36}-4=2\\ 6-4=2\\ 2=2$

x=16 is a solution

Then the solutions are x=0 and x=16

5 0

3 years ago

Given cos(O)=3/5 and tan(O)

Valentin [98]

Answer:

Tan(O) = 4/3

Step-by-step explanation:

According to SOH CAH TOA

cos theta = adj/hyp

cosO = 3/4

adj = 3

hyp = 5

Get the opposite

Using pythagoras theorem

opp^2 = hyp^2-adj^2

opp^2 = 5^2 - 3^2

opp^2 = 25-9

opp^2 = 16

opp = 4

Tan (O) = opp/adj

Tan(O) = 4/3

Hence Tan(O) = 4/3

3 0

2 years ago