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3 years ago

If two angles of a triangle measure 39° and 56°, what is the measure of the third angle? A. 85° B. 55° C. 105° D. 265°

Mathematics

1 answer:

yarga [219]3 years ago

8 0

Answer:

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

Subtract the sum of the 2 given angles from 180 for third angle

third angle = 180° - (39 + 56)° = 180° - 95° = 85° → A

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Enter the correct answer in the box. solve the equation x2 − 16x 54 = 0 by completing the square. fill in the values of a and b

poizon [28]

The roots of the given polynomials exist $x=8+\sqrt{10}$ , and $x=8-\sqrt{10}$ .

<h3>What is the formula of the quadratic equation?</h3>

For a quadratic equation of the form $a x^{2}+b x+c=0$ the solutions are

$x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Therefore by using the formula we have

$x^{2}-16 x+54=0$$

Let, a = 1, b = -16 and c = 54

Substitute the values in the above equation, and we get

$x_{1,2}=\frac{-(-16) \pm \sqrt{(-16)^{2}-4 \cdot 1 \cdot 54}}{2 \cdot 1}$$

simplifying the equation, we get

$$&x_{1,2}=\frac{-(-16) \pm 2 \sqrt{10}}{2 \cdot 1} \\$

$$&x_{1}=\frac{-(-16)+2 \sqrt{10}}{2 \cdot 1}, x_{2}=\frac{-(-16)-2 \sqrt{10}}{2 \cdot 1} \\$

$$&x=8+\sqrt{10}, x=8-\sqrt{10}$

Therefore, the roots of the given polynomials are $x=8+\sqrt{10}$ , and

$x=8-\sqrt{10}$ .

To learn more about quadratic equations refer to:

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3 0

1 year ago

Read 2 more answers

Please help using the picture below I need the surface area I also need an explanation

Taya2010 [7]

Answer:

155

Step-by-step explanation:

5*5=25

((5*13)/2)*4=130

130+25=155

6 0

2 years ago

1. Identify the DEPENDENT VARIABLE in the testable

MAVERICK [17]

To identify the dependent variable in the testable, look out for the variable that is affected by the other. The variable that changes as a result of another variable is the dependent variable.

In a research study, there are typically two main variables that direct the scientific enquiry. They are:

Dependent Variable, and
Independent Variable

The independent variable causes a change in the dependent variable, i.e. the dependent variable receives the effect, the independent variable is the cause of the change.

It is very easy to identify the dependent variable in any testable hypothesis once you are able to pick out which variable is causing a change in the other.

For example, let's say the topic of a research is: The Impact of Sunlight on Germination Rate of Seedlings.

Here, Sunlight is the independent variable affecting Germination Rate.

The dependent variable here would be: Germination Rate.

Therefore, to identify the dependent variable in the testable, look out for the variable that is affected by the other. The variable that changes as a result of another variable is the dependent variable.

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2 years ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago

Find the midpoint of MN M=-24 N=1

LekaFEV [45]

Midpoint=12.5 let me know if this helps

4 0

3 years ago