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emmainna [20.7K]
3 years ago
6

What values of x make 3 - 4x > 17 true?

Mathematics
1 answer:
julsineya [31]3 years ago
4 0

Answer: x < -3.5

Step-by-step explanation:

Subtract 3 from both sides and we get

-4x > 14

We divide -4 on both sides, but be careful because we have to flip the sign since it's a negative number.

x < -3.5

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Find the value of x.
nadya68 [22]

Answer:

127 degrees

Step-by-step explanation:

1. Find the angle next to "x" using the fact that all the angles in a triangle add to 180 degrees.

180-30-97 = 53

2. Angle X and the angle that is 53 degrees are supplementary meaning they will add to 180 degrees since it's on a line

180-53=57

Hope this helped :)

3 0
3 years ago
What is the measure of X! Please help if you know how to do this!
deff fn [24]

Answer:

28°

Step-by-step explanation:

You're given that line DE and line FG are parallel and KL and FG are perpendicular. Then you can find out angle ∠BAC by using the vertical angles property: ∠BAC=62°. Then since KL and FG are perpendicular ∠ABC = 90°. So you find the angle ∠BCA by finding the sum of interior angles: 62+90+∠BCA=180, therefore ∠BCA is 28°. Finally, ∠x or ∠JCG = 28 because ∠JCG and ∠BCA are vertical angles and congruent.

4 0
3 years ago
Translate the sentence into an equation.
maria [59]

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7 + (B÷5) = 3

Step-by-step explanation:

7 0
2 years ago
A bakery sold 246 vanilla cupcakes in a day, which was 82% of the total number of cupcakes sold that day. How many total cupcake
IRINA_888 [86]
The total number of cupcakes were 300.
5 0
3 years ago
Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"
vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}

which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
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