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bekas [8.4K]
3 years ago
5

Can y’all please help. I will send points if you want me to.

Mathematics
1 answer:
suter [353]3 years ago
8 0

Answer:

the third one has a 2 acute angles 2 abuse angles and the x value is 110

Step-by-step explanation

the x value has to be bigger then 90 degree angles because it is a abuts angle and the 57 angle is a acute angle

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A researcher at a large company has collected data on both the beginning salary and the current salary of 48 randomly selected e
Stolb23 [73]

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3 0
3 years ago
Work out the new Amount after a 10% increase<br><br> A) £14<br> B) 240g<br> C) 20p<br> D) 110mm
nasty-shy [4]

Answer:

a) £15.40 b) 260g c) 22p d) 121mm

Step-by-step explanation:

To work out a 10% increase, times the number by 1.10 or 1.1!

Hope this helped :)

3 0
3 years ago
The coordinates of the vertices of ∆PQR are P(-2,5), Q(-1,1), and R(7,3). Determine whether ∆PQR is a right triangle. Show your
Mashutka [201]

Given

∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

Determine whether ∆PQR is a right triangle

To proof

As given ∆PQR points are P(-2,5), Q(-1,1), and R(7,3)

First find out the sides of triangle

FORMULA

Distance formula between two points

D^{2}= (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}

 Distance   between two points P(-2,5) and Q(-1,1)

PR = \sqrt{(-1+2)^{2}+(1-5)^{2}  }

PR = \sqrt{17}

Distance between two points Q(-1,1)and  R(7,3)

QR = \sqrt{(7+1)^{2} +(3-1)^{2}  }

QR =\sqrt{68}

Distance between two points  R(7,3) and P(-2,5)

RP =\sqrt{(-2-7)^{2} + (5-3)^{2}  }

RP=\sqrt{85}

now show that ∆PQR is a right triangle

RP^{2} = PQ^{2} +QR^{2}

Putting the value given above

(\sqrt{85}) ^{2} = \sqrt{17} ^{2} +\sqrt{68} ^{2}

85 = 17 +68

85 =85

In the right triangle

HYPOTENUSE² = BASE² + PERPENDICULAR²

This is prove above

Hence ∆PQR is a right triangle

Hence proved










7 0
4 years ago
Solve the formula for h. A=bh
Leni [432]

Answer:

A/b = h

Step-by-step explanation:

A=bh

Divide each side by b

A/b=bh/b

A/b = h

3 0
4 years ago
Read 2 more answers
547321 rounded to nearest one thousand
Sphinxa [80]
The nearest thousand is 547,000
5 0
3 years ago
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