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VladimirAG [237]

3 years ago

6

A triangle has a side length of 4in in an area of 18 square inches and a larger similar triangle has a corresponding side length

of 8 in . Determine the area of the larger triangle

1 answer:

Harman [31]3 years ago

5 0

Answer:

72

<em>s1</em><em> </em><em>:</em><em>s2</em><em> </em><em>=</em><em> </em><em>A1</em><em> </em><em>:</em><em> </em><em>A2^</em><em>2</em>

<em>So</em><em> </em><em>the</em><em> </em><em>factor</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>sides</em><em> </em><em>is</em><em> </em><em>2</em><em>.</em><em> </em><em>Now</em><em> </em><em>you</em><em> </em><em>multiply</em><em> </em><em>the</em><em> </em><em>area</em><em> </em><em>by</em><em> </em><em>2</em><em> </em><em>squared</em><em>.</em>

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Quadrilateral ABCD is inscribed in this circle. What is the measure of angle C?

Shkiper50 [21]

Answer:

107 degrees

Step-by-step explanation:

For an inscribed quadrilateral, the opposite angles are supplementary. Thus, B+D = 180 and you can solve for x.

3x+9+2x-4 = 180

5x = 175

x = 35

Next, plug the x value into angle A.

2(35) + 3 = 73

Angle C is the supplement of this which is 180-73 = 107

6 0

3 years ago

Dose someone know how to figure this out can someone pleas help its a test

riadik2000 [5.3K]

Complete question is;

The levels of mercury in two different bodies of water are rising. In one body of water the initial measure of mercury is 0.05 parts per billion (ppb) and is rising at a rate of 0.1 ppb each year. In the second body of water the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Which equation can be used to find y, the year in which both bodies of water have the same amount of mercury?

A) 0.05 – 0.1y = 0.12 – 0.06y

B) 0.05y + 0.1 = 0.12y + 0.06

C) 0.05 + 0.1y = 0.12 + 0.06y

D) 0.05y – 0.1 = 0.12y – 0.06

Answer:

Option C: 0.05 + 0.1y = 0.12 + 0.06y

Step-by-step explanation:

In the first body, the initial measure of mercury is 0.05 parts per billion (ppb) while it's rising at a rate of 0.1 ppb each year. We are told to use y for the number of years.

Thus, amount of mercury for y years in this first body is;

A1 = 0.05 + 0.1y

Now, for the second body, we are told that;the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Thus, amount of mercury for y years in this second body is;

A2 = 0.12 + 0.06y

Since we want to find the year in which both bodies of water have the same amount of mercury. Thus, it means;

A1 = A2

Thus;

0.05 + 0.1y = 0.12 + 0.06y

7 0

3 years ago

Evaluate the numerical expression 4(8 - 2) = (-3) + 7

Fofino [41]

Answer:

4 ( 8 - 2 ) = (-3) + 7 -> this is a false statement

Step-by-step explanation:

4 ( 8 - 2 ) = ( -3 ) + 7

Simplify the parenthesis

4 ( 6 ) = (-3) + 7

Multiply;

24 = (-3) + 7

Add, remember, adding a negative number and a positive number is the same as subtracting the negative number from the positive number,

24 = 4

This statement is not true.

8 0

3 years ago

Read 2 more answers

A challenge to all math experts...

liq [111]

Answer:

0

Step-by-step explanation:

<h3>Given</h3>

(x + $\sqrt{1 + x^2}$ )(y + $\sqrt{1 + y^2}$ ) = 1
(x+y)² = ?

<h3>Solution</h3>

<u>Let's make substitution as:</u>

(x + $\sqrt{1 + x^2}$ ) = m

then

(y + $\sqrt{1 + y^2}$ ) = 1/m

<u>Solving the first equation for x and the second one for y:</u>

(x + $\sqrt{1 + x^2}$ ) = m
$\sqrt{1 + x^2}$ = m - x
1 + x² = (m - x)²
1 + x² = m² - 2mx + x²
2mx = m² - 1
x = (m² - 1)/2m

And

(y + $\sqrt{1 + y^2}$ ) = 1/m
$\sqrt{1 + y^2}$ = 1/m - y
1 + y² = 1/m² -2y/m + y²
1 = 1/m² - 2y/m
m² = 1 - 2my
2my = 1 - m²
y = (1 - m²)/2m
y = - (m² - 1)/2m

<u>Now, sum of x and y:</u>

x + y = (m² - 1)/2m - (m² - 1)/2m = 0

<u>Therefore:</u>

(x+y)² = 0

Answer is zero

8 0

3 years ago

The quadratic equation by completing the square x^2+6x=18

g100num [7]

This technique is valid only when the coefficient of x2 is 1.

1) Transpose the constant term to the right

x2 + 6x = −2.

<span> 2) <span>Add a square number to both sides -- add the square of half the coefficient of x. In this case, add the square of half of 6; that is, add the square of 3, which is 9:</span></span>

x2 + 6x + 9 = −2 + 9.

The left-hand side is now the perfect square of (x + 3).

(x + 3)2 = 7.

3 is half of the coefficient 6.

That equation has the form

<span><span><span>a2</span> = b</span> which implies<span>a = <span>±.</span></span> Therefore,<span><span>x + 3</span> = ±</span> <span>x = <span>−3 ±.</span></span></span>

That is, the solutions to

x2 + 6x + 2 = 0

3 0

3 years ago