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3 years ago

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Given the following diagram, if m∠1 = m∠4, then all of the following must be complementary angles except

2 answers:

olya-2409 [2.1K]3 years ago

8 0

<u>Answers</u>

angles 2 and 3

<u>Explanation</u>

From the diagram we have been told that <1 and <4 are equal.

Then <2 = <3 is true.

Complementary angles are two angles whose sum is 90°

<1 + <2 = 90°

angles 1 and 2 ⇒ They are complementary angles

<1 + <3 = 90°

angles 1 and 3 ⇒ They are complementary angles

<2 + <4 = 90°

angles 2 and 4 ⇒ They are complementary angles

angles 2 and 3 are equal but they are not complementary.

sukhopar [10]3 years ago

6 0

The answer is : angles 2 and 3

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Diana invested $3000 in a savings account for 3 years. She earned $450 in interest over that time period. What interest rate did

matrenka [14]

If I=P*r*t, we can input 450 in for I, 3000 in for the principal (P, or the starting rate), and 3 for t=time because there were three years, resulting in 450=9000*r. Next, we can divide both sides by 9000 to get 0.05 or 5% as our answer.

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3 years ago

Find all pairs of real numbers (a,b) such that (x-a)^2+(2x-b)^2=(x-3)^2+(2x)^2

Mrrafil [7]

Answer:

$(3,0)$

$(\frac{-9}{5},\frac{12}{5})$

Step-by-step explanation:

Let's expand both sides.

I'm going to use the following identity to expand the binomial squared expressions: $(u+v)^2=u^2+2uv+v^2$ or $(u-v)^2=u^2-2uv+v^2$ .

Left-hand side:

$(x-a)^2+(2x-b)^2$

$(x^2-2ax+a^2)+((2x)^2-2b(2x)+b^2)$

$x^2-2ax+a^2+4x^2-4bx+b^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-2ax-4bx)+(a^2+b^2)$

$5x^2+(-2a-4b)x+(a^2+b^2)$

Right-hand side:

$(x-3)^2+(2x)^2$

$x^2-2(3)x+9+4x^2$

$x^2-6x+9+4x^2$

Reorder so $x^2$ 's are together and that $x[tex]'s are together.[tex](x^2+4x^2)+(-6x)+9$

$5x^2-6x+9$

Now let's compare both sides.

If we want both sides to appear exactly the same we need to choose values $a$ and $b$ such the following are true equations:

$-2a-4b=-6$

$a^2+b^2=9$

So if we solve the system we can find the values $a$ and $b$ such that the left=right.

Let's solve the first equation for $a$ in terms of $b$ .

Add $2a$ on both sides:

$-4b=-6+2a$

Divide both sides by -4:

$b=\frac{-6+2a}{-4}$

Reduce (divide top and bottom by -2):

$b=\frac{3-a}{2}$

Now let's plug this into second equation:

$a^2+b^2=9$

$a^2+(\frac{3-a}{2})^2=9$

$a^2+\frac{9-6a+a^2}{4}=9$ (I used the identity $(u-v)^2=u^2-2uv+v^2$ )

Multiply both sides by 4 to clear the fractions from the problem:

$4a^2+(9-6a+a^2)=36$

Combine like terms on left hand side:

$4a^2+a^2-6a+9=36$

$5a^2-6a+9=36$

Subtract 36 on both sides:

$5a^2-6a-27=0$

Now let's try to factor.

We are going to try to find two numbers that multiply to be 5(-27) and add to be -6.

5(-27)=(5*3)(-9)=15(-9)=-15(9) while -15+9=-6.

So let's replace $-6a$ with $-15a+9a$ and factor by grouping.

$5a^2-15a+9a-27=0$

$5a(a-3)+9(a-3)=0$

$(a-3)(5a+9)=0$

This implies $a-3=0$ or $5a+9=0$ .

Solving the first is easy. Just ad 3 on both sides to get: $a=3$ .

The second requires two steps. Subtract 9 and then divide by 5 on both sides.

$5a=-9$

$a=\frac{-9}{5}$ .

So let's go back to finding $b$ now that we know the $a$ values.

If $a=3$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-3}{2}=0$ .

So one ordered pair $(a,b)$ that satisfies the equation is:

$(3,0)$ .

If $a=\frac{-9}{5}$ and $b=\frac{3-a}{2}$ ,

then $b=\frac{3-\frac{-9}{5}}{2}$ .

Let's multiply top and bottom by 5 to clear the mini-fraction.

$b=\frac{15-(-9)}{10}$

$b=\frac{24}{10}$

$b=\frac{12}{5}$

So one ordered pair $(a,b)$ that satisfies the equation is:

$(\frac{-9}{5},\frac{12}{5})$ .

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4 years ago

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4 years ago

Alexandra bought a $90000 life insurance policy at $10.98 for a 20 year term. She will pay ___ over 20 years for the premium

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4 years ago

Which ordered pairs make the equation true?

Angelina_Jolie [31]

Answer:

(-3, 1) and (2, -1)

Step-by-step explanation:

We'll just substitute in these coordinates.

<u>First Pair:</u>

2(-3) + 5(1) = -1

-6 + 5 = -1

-1 = -1

This works!

<u>Second Pair:</u>

2(5) + 5(-3) = -1

10 - 15 = -1

-5 ≠ -1

This does not work.

<u>Third Pair:</u>

2(2) + 5(-1) = -1

4 - 5 = -1

-1 = -1

This also works!

<u>Fourth Pair:</u>

2(-4) + 5(2) = -1

-8 + 10 = -1

2 ≠ -1

This does not work.

6 0

3 years ago