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4 years ago

How many integers are there between -8 and 2? Group of answer choices

Mathematics

1 answer:

Dahasolnce [82]4 years ago

8 0

Answer:

Step-by-step explanation:

{-7,-6,-5,-4,-3,-2,-1,0,1 }

9 numbers between -8 and 2.

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Please help 10 minutes left

Maksim231197 [3]

Answer:

I belive the answer is D

Step-by-step explanation:

6 0

3 years ago

A landmark on the first map is a triangle with side lengths of 3 cm, 4 cm, and 5 cm. What are the side lengths of the triangle l

Sveta_85 [38]

Answer:

the side lengths of the second triangle are 2.25mm ,3m, and 3.75mm

the work shown:

3mm 3mm × 3cm/4cm = 2.25mm

4mm 4mm × 3cm/4cm = 3mm

5mm 5mm × 3cm/4cm = 3.75mm

3 0

3 years ago

Convert the angle 8pi/9 radians to degrees<br><br>express your answer exactly

Thepotemich [5.8K]

Answer:

hello : 160°

Step-by-step explanation:

8pi/9 = (8×180°)/9 = 160°

6 0

3 years ago

Read 2 more answers

Solve the formula for the indicated variable. x-2/2 = m + n for x

never [62]

Given equation as
$\frac{x-2}{2} = m+n$

In this we have to isolate x. So for this firstly we will multiply both side of equation by 2.
$\frac{(x-2)}{2} * 2 = (m+n)*2$
$(x-2) = 2(m+n)$

Now we can add 2 in both side
$x-2+2 = 2(m+n) + 2$
$x = 2(m+n) + 2$

8 0

3 years ago

Read 2 more answers

The Segment joining the midpoints of sides PQ and PR (the midsegment) in traingle PQR has length 8x^12x+36. If the length of QR

Nata [24]

Answer:

The possible measures of the midsegment are <u>32 units and 212 units</u>.

Step-by-step explanation:

Given:

A triangle PQR with the midsegment made by sides PQ and PR has length equal to $8x^2+12x+36$ and the base length QR opposite to the midsegment is $4x^2+60x+120$ .

From midsegment theorem, we know that, the midsegment is a line parallel to the base opposite to it and half the length of the base length.

Therefore, Midsegment = $\frac{1}{2}\times$ Base length QR

$8x^2+12x+36=\frac{1}{2}(4x^2+60x+120)\\\\8x^2+12x+36=2x^2+30x+60\\\\(8x^2-2x^2)+(12x-30x)+(36-60)=0\\\\6x^2-18x-24=0\\\\6(x^2-3x-4)=0\\\\x^2-3x-4=0\\\\x^2+x-4x-4=0\\\\x(x+1)-4(x+1)=0\\\\(x+1)(x-4)=0\\\\x=-1\ or\ x=4$

Now, midsegment can be calculated using the values of 'x'.

First, plug in -1 for 'x'. This gives,

$Midsegment=8(-1)^2+12(-1)+36\\\\Midsegment=8\times 1-12+36\\\\Midsegment=8-12+36=32\ units$

Now, plug in 4 for 'x'. This gives,

$Midsegment=8(4)^2+12(4)+36\\\\Midsegment=8\times 16-48+36\\\\Midsegment=128+48+36=212\ units$

Therefore, the possible measures of the midsegment are 32 units and 212 units

3 0

3 years ago