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2 years ago

13

Triangle Angle Sum Theorem

1 answer:

Kipish [7]2 years ago

7 0

Answer:

m<DCF = 155°

Step-by-step explanation:

m<DCF = m<D + m<E

m<DCF = 72° + 83°

m<DCF = 155°

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What is the equation of the line in the following graph? 1)y=x+2

Levart [38]

The slope-intercept form:

$y=mx+b$

m - slope

b - y-intercept

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have two points (2, 0) and (-2, -4). Substitute:

$m=\dfrac{-4-0}{-2-2}=\dfrac{-4}{-4}=1$

Therefore we have the equation of a line

$y=1x+b\to y=x+b$

Put the coordinates of the point (2, 0) to the equation:

$0=2+b$ <em>subtract 2 from both sides</em>

$-2=b\to b=-2$

Answer: $\boxed{y=x-2}\to\boxed{3)}$

8 0

4 years ago

Arachnophobia is the fear of spiders. People with arachnophobia suffer to different degrees, but they tend to react intensely to

zepelin [54]

Answer:

1. The null hypothesis is that there is no difference between the population means (in other words, there is no treatment effect).

2. The<u> alternative hypothesis</u> is that at least one of the population means is different from another (in other words, there is an effect Of at least one Of the treatments).

Step-by-step explanation:

In a statistical analysis, the statistician shares his participants into several groups and administers treatments to them. After administering the treatment, he might want to compare the values obtained to see if there is a difference. If he accepts the null hypothesis as true, that means that there was no difference between the population means. This can also be taken to mean that the result had no difference among the groups.

But if the alternative hypothesis was true, that means that the population means were not all equal and so, there was an effect from the treatment administered.

7 0

3 years ago

PLEASE HELP. it is in the attachment

juin [17]

Answer:

<JKL=123

Step-by-step explanation:

57+57=114

360-114=246

246/2=123

<JKL=123

3 0

3 years ago

Read 2 more answers

PLEASE HELP! Which ordered pairs are solutions to the inequality y-4x (greater than or equal to) -5? Select each correct answer

Wittaler [7]

(-2, 1), (-4, 2) and (1, -1) are solutions to given inequality y-4x (greater than or equal to) -5

<u>Solution:</u>

Given, inequality is y – 4x ≥ - 5

We have to find the correct answers from given options

<em><u>We have to check option by option. </u></em>

Now, (-2, 1) ⇒ 1 – 4(-2) ≥ - 5 ⇒ 1 + 8 ≥ -5 ⇒ 9 ≥ -5 ⇒ correct answer.

Now, (-4, 2) ⇒ 2 – 4(-4) ≥ -5 ⇒ 2 + 16 ≥ - 5 ⇒ 18 ≥ - 5 ⇒ correct answer

Now, (1, -1) ⇒ -1 – 4(1) ≥ - 5 ⇒ -1 – 4 ≥ - 5 ⇒ -5 ≥ -5 ⇒ correct answer

Now, (4, 0) ⇒ 0 – 4(4) ≥ - 5 ⇒ -16 ≥ -5 ⇒ wrong answer

Now, (5, -2) ⇒ -2 – 4(5) ≥ - 5 ⇒ -2 – 20 ≥ -5 ⇒ -22 ≥ -5 ⇒ wrong answer

Hence, (-2, 1), (-4, 2) and (1, -1) are solutions to given inequality.

8 0

3 years ago

What is the largest possible integral value in the domain of the real-valued function

kotegsom [21]

Answer:

Max Value: x = 400

General Formulas and Concepts:

<u>Algebra I</u>

Domain is the set of x-values that can be inputted into function f(x)

<u>Calculus</u>

Antiderivatives
Integral Property: $\int {cf(x)} \, dx = c\int {f(x)} \, dx$
Integration Method: U-Substitution
[Integration] Reverse Power Rule: $\int {x^n} \, dx = \frac{x^{n+1}}{n+1} + C$

Step-by-step explanation:

<u>Step 1: Define</u>

$f(x) = \frac{1}{\sqrt{800-2x} }$

<u>Step 2: Identify Variables</u>

<em>Using U-Substitution, we set variables in order to integrate.</em>

$u = 800-2x\\du = -2dx$

<u>Step 3: Integrate</u>

Define: $\int {f(x)} \, dx$
Substitute: $\int {\frac{1}{\sqrt{800-2x} } } \, dx$
[Integral] Int Property: $-\frac{1}{2} \int {\frac{-2}{\sqrt{800-2x} } } \, dx$
[Integral] U-Sub: $-\frac{1}{2} \int {\frac{1}{\sqrt{u} } } \, du$
[Integral] Rewrite: $-\frac{1}{2} \int {u^{-\frac{1}{2} }} \, du$
[Integral - Evaluate] Reverse Power Rule: $-\frac{1}{2}(2\sqrt{u}) + C$
Simplify: $-\sqrt{u} + C$
Back-Substitute: $-\sqrt{800-2x} + C$
Factor: $-\sqrt{-2(x - 400)} + C$

<u>Step 4: Identify Domain</u>

We know from a real number line that we cannot have imaginary numbers. Therefore, we cannot have any negatives under the square root.

Our domain for our integrated function would then have to be (-∞, 400]. Anything past 400 would give us an imaginary number.

7 0

3 years ago