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motikmotik
3 years ago
12

What is the midpoint of AB?

Mathematics
1 answer:
ella [17]3 years ago
3 0

Answer:

(4, 0.5)

Step-by-step explanation:

l

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Length of a rectangle is 5 cm longer than the width. Four squares are constructed outside the rectangle such that each of the sq
TiliK225 [7]

Answer:

The perimeter of rectangle is 18\ cm

Step-by-step explanation:

Let

x-----> the length of the rectangle

y----> the width of the rectangle

we know that

x=y+5 ----> equation A

120=xy+2x^{2}+2y^{2}  ---> equation B (area of the constructed figure)

substitute the equation A in equation B

120=(y+5)y+2(y+5)^{2}+2y^{2}

120=(y+5)y+2(y+5)^{2}+2y^{2}\\ 120=y^{2}+5y+2(y^{2}+10y+25)+2y^{2}\\ 120=y^{2}+5y+2y^{2}+20y+50+2y^{2}\\120=5y^{2}+25y+50\\5y^{2}+25y-70=0

using a graphing calculator -----> solve the quadratic equation

The solution is

y=2\ cm

Find the value of x

x=y+5 ----> x=2+5=7\ cm

Find the perimeter of rectangle

P=2(x+y)=2(7+2)=18\ cm

4 0
3 years ago
Read 2 more answers
PLEASE HELP!!!!!!!!!!!!1
skelet666 [1.2K]
A. Let x = cheese and
y = chocolate
2x + y = 25
x + y = 20

B. Subtract the second equation from the first.
2x + y = 25
-(x + y = 20)
-—————
x = 5

Plug 5 back in to the second equation and solve for y.
x + y = 20
5 + y = 20
Subtract 5 from both sides.
y = 15
5 cheese and 15 chocolate

Used elimination method because coefficients on the y values were both 1 so it was easy to subtract the equations and eliminate the y variable.


5 0
3 years ago
2x² - 5x+1 has roots alpha and beta. Find alpha⁴+beta⁴ without solving the equation.
shepuryov [24]

Answer:

Step-by-step explanation:

\alpha+\beta=\dfrac{5}{2} \\\\\alpha*\beta=\dfrac{1}{2} \\\\\alpha^2+\beta^2=\dfrac{21}{4} \ (see\ previous\ post)\\\\(\alpha+\beta)^4=\dfrac{625}{16} \\\\=\alpha^4+\beta^4+4*\alpha^3*\beta+6*\alpha^2*\beta^2+4*\alpha*\beta^3\\\\=\alpha^4+\beta^4+4*(\alpha*\beta)(\alpha^2+\beta^2)+6*\alpha^2*\beta^2\\\\\alpha^4+\beta^4=(\alpha+\beta)^4-4*(\alpha*\beta)(\alpha^2+\beta^2)-6*\alpha^2*\beta^2\\\\= \dfrac{625}{16} -4*\dfrac{1}{2} *\dfrac{21}{4} -6*(\dfrac{1}{2})^2 \\\\

= \dfrac{625}{16}- \dfrac{168}{16}-\dfrac{24}{16}\\\\\\= \dfrac{433}{16}

7 0
3 years ago
A boat is 60m from the base of a lighthouse. The angle of depression between the lighthouse and the boat is 37°. How tall is the
LekaFEV [45]

Answer: 34.64 m

Step-by-step explanation:

Given: A boat is 60 m from the base of a lighthouse.

The angle of depression between the lighthouse and the boat is 37°.

By using trigonometric ratios :

\tan x=\dfrac{\text{Side opposite to }x}{\text{Side adjacent to }x}

here x= 37°, side opposite to x = height of lighthouse (h) , side adjacent to x = 60 m

\tan 37^{\circ}=\dfrac{h}{60}\\\\\Rightarrow\ 0.57735=\dfrac{h}{60}\\\\\Rightarrow\ h= 60\times0.57735\approx34.64

Hence, the lighthouse is 34.64 m tall.

6 0
3 years ago
The diameters of ball bearings are distributed normally. The mean diameter is 83 millimeters and the standard deviation is 3 mil
Alenkinab [10]

Answer:

0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 83, \sigma = 3

Find the probability that the diameter of a selected bearing is greater than 85 millimeters.

This is 1 subtracted by the pvalue of Z when X = 85. Then

Z = \frac{X - \mu}{\sigma}

Z = \frac{85 - 83}{3}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486.

1 - 0.7486 = 0.2514

0.2514 = 25.14% probability that the diameter of a selected bearing is greater than 85 millimeters.

7 0
3 years ago
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