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DanielleElmas [232]

4 years ago

6

Math multiple choice question.

1 answer:

bekas [8.4K]4 years ago

3 0

Infinite solutipn is the answet

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Will give branliest to first person! please help: asap!!!!!!!!!!!

BartSMP [9]

What I would do first is find the base and height of the rectangle and from there you have the diameter of the circle so you can divide that by 2 and you have your radius which you’d be able to find the total area

4 0

4 years ago

Can anyone please help me with this?

AnnyKZ [126]

Answer:

m∠QPT = 110°

Step-by-step explanation:

1. We can use exterior angle theorem to find m∠QPT. Exterior Angle Theorem states that the exterior angle of a triangle is equal to the sum of the two remote angles in that same triangle.

This means that 6x + 20 = x + 40 + 4x - 5. Then, we can plug in the value of x into 6x + 20 and find the measure of ∠QPT.

2. (Solving)

Step 1: Combine like terms.

$6x + 20 = (x+4x)+(40-5)$
$6x + 20 = 5x + 35$

Step 2: Subtract 20 from both sides.

$6x + 20 - 20 = 5x + 35 - 20$
$6x = 5x +15$

Step 3: Subtract 5x from both sides.

$6x - 5x = 5x +15 - 5x$
$x = 15$

Step 4: Check if solution is correct.

$6(15)+20=(15)+40+4(15)-5$
$90 + 20 = 15 + 40 + 60 - 5$
$110 = 55 + 60 - 5$
$110 = 115 - 5$
$110 = 110$

3. Now, let's plug in 15 for 6x + 20:

$6(15)+20$

$90+20$
$110$

Therefore, m∠QPT = 110°.

7 0

3 years ago

Read 2 more answers

What is the solution set of x2 + 5x- 5 = 0? oss-355 S +385 o sus S* o {stay -S-345 , 30 O S=5+375 5+3/5 2 0 --375 -5 +385 2

jolli1 [7]

Given the quadratic equation:

$x^2+5x-5=0$

Using the general solution for quadratic equations:

$ax^2+bx+c=0\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

From the problem, we identify:

$\begin{gathered} a=1 \\ b=5 \\ c=-5 \end{gathered}$

Then:

$x=\frac{-5\pm\sqrt{25+20}}{2}=\frac{-5\pm3\sqrt{5}}{2}$

And the solution set is:

$\lbrack\frac{-5-3\sqrt{5}}{2},\frac{-5+3\sqrt{5}}{2}]$

6 0

1 year ago

Given the equation y=14x2−5, the minimum is ___ and the range is ___

mel-nik [20]

Answer:

The minimum is -5

The range is $y \ge -5$

f(14) = 2739

Step-by-step explanation:

Given

$y = 14x^2 - 5$

Solving (a): The minimum

A quadratic function is represented as:

$y = ax^2 + bx + c$

If a > 0, then the function has a minimum

By comparison

$a = 14$ --- the function has a minimum

$b = 0$

$c = -5$

To calculate the minimum, we first calculate the following is calculated as:

$m = -\frac{b}{2a}$

So, we have:

$m = -\frac{0}{2*14}$

$m = 0$

So, the minimum is at f(m)

We have: $y = 14x^2 - 5$

$f(0) = 14 *0^2 - 5$

$f(0) = - 5$

Solving (b): The range

In (a), we have:

$f(0) = - 5$ --- the minimum

This implies that the smallest value of y on the graph is -5.

So, the range is:

$r = \{y|y\ge -5\}$

Solving (c): f(14)

We have:

$y = 14x^2 - 5$

So:

$f(14) = 14 * 14^2 - 5$

$f(14) = 2739$

4 0

3 years ago

Which of the following pairs of polnts are on the same horizontal line?

aalyn [17]

Answer:

The answer is A. (-2,-2 and (2,2)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers