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2 years ago

Given ABC = DEF, m DE = 5x - 4. Find the value of x.?

Mathematics

1 answer:

lesantik [10]2 years ago

7 0

There something you ain’t giving us in this equation

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George jogged a total distance of 9 and 1 over 2 miles during the months of October and November. If George only jogged 1 over 4

antiseptic1488 [7]

Hello,

Question:

George jogged a total distance of 9 and 1 over 2 miles during the months of October and November. If George only jogged 1 over 4 mile every day, which expression shows the number of days in which he went jogging?

We know:

He jogged 9 1/2 miles during October and November

1/4 mile every day

Answer:

9 and 1 over 2 divided by 1 over 4

6 0

2 years ago

Read 2 more answers

Simplify the expression, 728x6−−−−√−563x6−−−−√ . Assume all variables are positive.

Karo-lina-s [1.5K]

Answer: $-3x\sqrt{7}$

Step-by-step explanation:

Don't have much of one, but just took test

3 0

2 years ago

Staph the inequality and plot a point in the solution set. Explain how you know the point is a solution.

seropon [69]

Answer:

The points when graphed are: (3,0) and (0,-5)

Step-by-step explanation:

I know that these points are a solution to the graph because it has one x value and one y value that stays constant.

For an example, if we used an X and Y table we could see that each variable has one other variable.

This is how it look like when graphed:

8 0

3 years ago

Any one Know the Answer To This Algebra 1

alexdok [17]

Answer:

The correct answer is B. -3.

Step-by-step explanation:

To solve this problem, we need to plug in -3 for all of the values of x in the equation. If we do this, we get:

f(x) = 2x + 3

f(-3) = 2(-3) + 3

Now, we can just simplify the expression found on the right side of the equation using PEMDAS. First, we should perform the multiplication.

f(-3) = -6 + 3

Next, we should add our final two terms together.

f(-3) = -3

Therefore, the correct answer is -3.

Hope this helps!

3 0

3 years ago

Tan 3A in terms of tan

Zanzabum

Here's the sum rule for the tangent function:

$\tan(a+b)=\dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$

In the special case a=b, this becomes the double angle formula:

$\tan(a+a)=\tan(2a)=\dfrac{\tan(a)+\tan(a)}{1-\tan(a)\tan(a)}=\dfrac{2\tan(a)}{1-\tan^2(a)}$

In your case, you case use the sum rule once:

$\tan(3a)=\tan(2a+a)=\dfrac{\tan(2a)+\tan(a)}{1-\tan(2a)\tan(a)}$

And use it again, in the special case of the double angle:

$\dfrac{\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a)}{1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a)}$

We can obvisouly simplify this expression a lot: let's deal with the numerator and denominator separately: the numerator is

$\dfrac{2\tan(a)}{1-\tan^2(a)}+\tan(a) = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}$

and the denominator is

$1-\dfrac{2\tan(a)}{1-\tan^2(a)}\tan(a) = \dfrac{1-\tan^2(a)-2\tan^2(a)}{1-\tan^2(a)} = \dfrac{1-3\tan^2(a)}{1-\tan^2(a)}$

So, the fraction is

$\dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-\tan^2(a)}\cdot \dfrac{1-\tan^2(a)}{1-3\tan^2(a)} = \dfrac{2\tan(a)+\tan(a)-\tan^3(a)}{1-3\tan^2(a)}$

8 0

3 years ago