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2 years ago

Match the inequalities with their solutions i just need the equation

Mathematics

2 answers:

storchak [24]2 years ago

5 0

6x>0 is the answer for the second one on the bottom
3x-4<1 is the answer for the first one on the bottom
4x? 2 I don’t know what sign above would be

Nataliya [291]2 years ago

4 0

JqhdifbeukwhskzbdhffibezhGaisvdcsbsgidbdhdudbegdzudbzudjddjdjdj

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Andrew has three times as many pencils as Noah and five pens he has 17 pens and pencils altogether how many pencils does Noah ha

Novay_Z [31]

9514 1404 393

Answer:

Step-by-step explanation:

The distributive property does not appear to be involved in this question.

Let n represent the number of Noah's pencils. Then Andrew's total is ...

3n +5 = 17

3n = 12 . . . . . subtract 5 from both sides

n = 4 . . . . . . divide both sides by 3

Noah has 4 pencils.

7 0

2 years ago

The shape above has the following

Ira Lisetskai [31]

It’s A 44-1) and if not then it’s b

8 0

2 years ago

40 pts and Brainliest pls give detailed answer :)

horsena [70]

Answer:

58 ft

Step-by-step explanation:

So I attached a diagram that illustrates the triangle that is formed. We know an angle, as well as the hypotenuse. We are looking for the height, or in other words the opposite side of the angle. There is a trigonometric function defined as: $sin(\theta) = \frac{opposite}{hypotenuse}$ . Using this we can plug in known values and solve for the opposite side, which I'll simply represent as x.

$sin(46) = \frac{x}{80}$

Multiply both sides by 80

$sin(46) * 80 = x$

Calculate sin(46) using a calculator (make sure it's in degree mode)

$0.7193398 * 80 = x$

Simplify

$57.547 \approx x$

Round this to the nearest foot

$58 \approx x$

4 0

1 year ago

How do I simplify the n^-6*n^3 topic "Negative and Zero Exponents"

Alchen [17]

Since you’re multiplying n^-6 and n^3, you can add the exponents: -6+3 = -3

n^-6 * n^3 = n^-3

If you need to finish without having a negative exponent in your answer, then remember a negative exponent means that factor is on the wrong side of the fraction.

n^-3 = n^-3 / 1 = 1/n^3

when the factor moves to the other side of the fraction, the side of the exponent changes.

7 0

2 years ago

2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

<u>Slope:</u>

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

<u>Slope:</u>

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$

So the diagonals measure the same, therefore this is a rectangle.

5 0

3 years ago