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2 years ago

Using the order of operation which should be done first to evaluate 6+(-5-7)/2-8(3)?

Mathematics

1 answer:

valentinak56 [21]2 years ago

6 0

Answer:

$\frac{3}{11}$

Step-by-step explanation:

=> $\frac{6+(-5-7)}{2-8(3)}$

=> $\frac{6+(-12)}{2-24}$

=> $\frac{6-12}{-22}$

=> $\frac{-6}{-22}$

=> $\frac{3}{11}$

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Please help me out with this so I can keep my kitten.

notka56 [123]

Answer:

$y = \frac{32}{3}$

Step-by-step explanation:

Firstly, move over the negative 3/4 fraction (don't forget to swap the operation i.e subtract to add):

$\frac{y}{8} = \frac{7}{12} + \frac{3}{4}$

Now, to add the two fractions, simply multiply the numerator and denominator by 3:

$\frac{3*3}{4*3} = \frac{9}{12}$

Now add this to the other fraction:

$\frac{9}{12} + \frac{7}{12} = \frac{16}{12}$

This can be simplified down by dividing both the numerator and denominator by 4:

$\frac{4}{3}$

Which now simplifies the original equation to:

$\frac{y}{8} = \frac{4}{3}$

Remove the y out of the fraction:

$\frac{1}{8}y = \frac{4}{3}$

Now multiply both sides by 8:

$(\frac{1}{8}y) * 8 = (\frac{4}{3}) * 8$

$y = \frac{4*8}{3}$

$y = \frac{32}{3}$

Hope this helps!

5 0

3 years ago

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A line passes through the two given points. Is it vertical, horizontal,or neither?<br>(8,3), (8,-4)

Katena32 [7]

Answer:

horizontal.

Step-by-step explanation:

it goes straight up

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3 years ago

The height of a trapezoid is 6 in. and it’s area is 72 in to the 2nd power. One base of the trapezoid is 6 inches longer than th

Vitek1552 [10]

Given:

The height of the given trapezoid = 6 in

The area of the trapezoid = 72 in²

Also given, one base of the trapezoid is 6 inches longer than the other base

To find the lengths of the bases.

Formula

The area of the trapezoid is

$A=\frac{1}{2} (b_{1} +b_{2} )h$

where, h be the height of the trapezoid

$b_{1}$ be the shorter base

$b_{2}$ be the longer base

As per the given problem,

$b_{2}=b_{1} +6$

Now,

Putting, A=72, $b_{2}=b_{1}+6$ and h=6 we get,

$\frac{1}{2} (b_{1} +b_{1}+6)(6) = 72$

or, $b_{1}+b_{1}+6 = \frac{(72)(2)}{6}$

or, $2b_{1}+6 = 24$

or, $2b_{1}=24-6$

or, $b_{1}= \frac{18}{2}$

or, $b_{1}=9$

So,

The shorter base is 9 in and the other base is = (6+9) = 15 in

Hence,

One base is 9 inches for one of the bases and 15 inches for the other base.

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3 years ago

SOMEBODY PLEASE HELP QUICK!!! WILL GIVE BRAINLIEST

Anna71 [15]

Carlos is correct

Since we don't know the length of sides PR and XZ, the triangles can't be congruent by the SSS theorem or the SAS theorem, and since we don't know the measure of angles Y and Q, the triangles can't be congruent by the ASA theorem, the SAS theorem or the AAS theorem. Therefore, Carlos is correct.

Carlos is correct. Since the angles P and X are not included between PQ and RQ and XY and YZ, the SAS postulate cannot be used, since it states that the angle must be included between the sides. Unlike with ASA, where there is the AAS theorem for non-included sides, there is not SSA theorem for non-included angles, so the triangles cannot be proven to be congruent.

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3 years ago

How do you use the distance formula and slope formula to classify a triangle

Sergio039 [100]

see the distance formula to find the length of the sides...
opposite sides equal it could be a rectangle or parallelogram
all sides equal, square or rhombus
adjacent equal, kite
and then the slope is used to check angles
if the product of the 2 lines in -1 the lines are perpendicular (right angle)
the the slopes of 2 lines are the same the sides are parallel.
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3 years ago

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