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3 years ago

Help me in geometry 100% right only

Mathematics

2 answers:

Marysya12 [62]3 years ago

5 0

C) C, A, B is the correct answer.

dimulka [17.4K]3 years ago

3 0

The correct answer is C) <C, <A, <B. You can tell because the angle opposite the shortest side is the smallest angle, the angle opposite the longest side is the largest angle, etc.

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Which of these is NOT equivalent to 2-^3.

shusha [124]

2^2 times 2^-5 i think but i am not positive

7 0

3 years ago

4(3x + 2) =16x + 7 -4x + 1

GrogVix [38]

Answer:

All real numbers are solutions.

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

4(3x+2)=16x+7−4x+1

(4)(3x)+(4)(2)=16x+7+−4x+1(Distribute)

12x+8=16x+7+−4x+1

12x+8=(16x+−4x)+(7+1)(Combine Like Terms)

12x+8=12x+8

Step 2: Subtract 12x from both sides.

12x+8−12x=12x+8−12x

8=8

Step 3: Subtract 8 from both sides.

8−8=8−8

0=0

7 0

3 years ago

Read 2 more answers

What is the sum of the last 5 terms of the series 1 9 17... 49?

drek231 [11]

The series is increasing by 8 each time.

So that means the complete series is:
1, 9, 17, 25, 33, 41, 49

Add the last 5 digits:
49 + 41 + 33 + 25 + 17 = 165

Answer:
165

4 0

4 years ago

Read 2 more answers

The line $y = 3$ intersects the graph of $y = 4x^2 + x - 1$ at the points $A$ and $B$. The distance between $A$ and $B$ can be w

raketka [301]

Answer:

Step-by-step explanation:

Let's find the points $A$ and $B$ .

We know that the $y$ -coordinates of both are $3$ .

So let's first solve:

$3=4x^2+x-1$

Subtract 3 on both sides:

$0=4x^2+x-1-3$

Simplify:

$0=4x^2+x-4$

I'm going to use the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , to solve.

We must first compare to the quadratic equation, $ax^2+bx+c=0$ .

$a=4$

$b=1$

$c=-4$

$\frac{-1 \pm \sqrt{1^2-4(4)(-4)}}{2(4)}$

$\frac{-1 \pm \sqrt{1+64}}{8}$

$\frac{-1 \pm \sqrt{65}}{8}$

Since the distance between the points $A$ and $B$ is horizontal. We know this because they share the same $y-coordinate$ .This means we just need to find the positive difference between the $x$ -values we found for the points of $A$ and $B$ .