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snow_tiger [21]

4 years ago

8

I need help please! Ashton estimates the square root of 72 in the following way: square root 72 = 2square root 36 = 2 times 6 =

12 (a) explain why his reasoning is incorrect. (B) estimate the square root of 72 to the nearest tenth without using your calculator

1 answer:

timofeeve [1]4 years ago

5 0

Answer:

B. Estimate the square root of 72 to the nearest tenth without using your calculator

Step-by-step explanation:

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yaroslaw [1]

I dont see any steps.

4 0

3 years ago

Read 2 more answers

How many solutions are there to the equation below? 17 x - 8 = 3x + 16<br>maqth

vazorg [7]

Answer:

There is one solution and that answer is mixed number form 1 5/7 or exact form 12/7

Step-by-step explanation:

8 0

3 years ago

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

I need help what’s the answer ?

pogonyaev

Answer:

x = 30

Step-by-step explanation:

Using the sine ratio in the right triangle

sinx = $\frac{opposite}{hypotenuse}$ = $\frac{2.5}{5}$ = $\frac{1}{2}$ , then

x = $sin^{-1}$ ( $\frac{1}{2}$ ) = 30

4 0

3 years ago

The equation of a circle centered at the origin is x^2+y^2=16. what is the radius of the circle?

vekshin1

Answer:

The center is (0,0) and the radius is 4

Step-by-step explanation:

x^2+y^2=16.

The equation of a circle can be written in the form

(x-h)^2+(y-k)^2=r^2 where ( h,k) is the center and r is the radius

(x-0)^2+(y-0)^2=4^2

The center is (0,0) and the radius is 4

4 0

3 years ago