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3 years ago

Can someone help me wit da question listed in da picture?

Mathematics

2 answers:

satela [25.4K]3 years ago

7 0

Answer:

it would be 14x

Step-by-step explanation:

ycow [4]3 years ago

5 0

Answer:

Option C)

Step-by-step explanation:

When Solving these equations rember PEMDAS

Parentheses
Exponits
Multiply
Divide
Add
Subtract

Hope this helps, PEMDAS is just an easy way to show what to do first, second, Ect.

Please Rate Brainiest!

-Aslina

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The following frequency table shows the number of cavities each of Dr. Freese's patients had.

Anit [1.1K]

Answer:

Step-by-step explanation:

13 patients had 0 cavities, according to the table! The righthand column is the frequency of people, and the left is the number of cavities they had.

6 0

3 years ago

25h + 40 = −15h − 80

elena-s [515]

Answer:

Step-by-step explanation:

25h + 40 = −15h − 80

25h + 15h = -80 - 40

40h = - 120

h = -120/4

h= -3

4 0

3 years ago

Read 2 more answers

Use the circle to solve for the missing measures. Assume lines that appear tangent are tangent. Do NOT round any answers.

wlad13 [49]

The given circle having two external tangents, according to circle

theorem, have the following values;

r = 3

m∠1 = 82°

m∠2 = 49°

x = 82°

<h3>How can the radial length and angles be found?</h3>

r² + 1.6² = (0.4 + r)²

(0.4 + r)² - (r² + 1.6²) = 0

0.8·r - 2.4 = 0

0.8·r = 2.4

r = 2.4 ÷ 0.8 = 3

<u>r = 3</u>

y = 49°

From the two tangent angle theorem, we have;

m∠1 = ((82 + 98 + 82) - (360 - ((82 + 98 + 82)))) ÷ 2 = 82

<u>m∠1 = 82°</u>

According to circle theorems, angle at the center is twice angle

subtended at the circumference.

m∠2 = (360 - (82 + 98 + 82)) ÷ 2 = 49°

<u>m∠2 = 49°</u>

The length of the chord subtended by the arc <em>x</em> is equal to the length of

the chord subtended by 89°, which gives;

<u>x = 82°</u>

Learn more about circle theorems applications here:

brainly.com/question/16879446

4 0

2 years ago

Ex 2.8<br> 3. find the maximum value of y for the curve y=x^5 -3 for -2≤x≤1

harkovskaia [24]

$y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\
y''(0)=20\cdot0^3=0$

The value of the second derivative for $x=0$ is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of $5x^4$ is always positive for $x\in\mathbb{R}\setminus \{0\}$ . That means at $x=0$ there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval $[-2,1]$ .
The function $y$ is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.

$y_{max}=y(1)=1^5-3=-2$

4 0

3 years ago

Vega wants to prove that the segment joining midpoints of two sides of a triangle is half the length of the third side. Select t

konstantin123 [22]

Answer:

The Answer is D : In triangle EFG△ , if EH=HF and EI=IG , then HI= 1/2 FG

FGH, I, equals, start fraction, 1, divided by, 2, end fraction, F, G.

Step-by-step explanation:

I Saw on Khan Academy

7 0

3 years ago