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vaieri [72.5K]
3 years ago
7

Find x, y, z please help!

Mathematics
1 answer:
ki77a [65]3 years ago
6 0

Answer:

x is 1 y is 1 and a is 1 hope this helped

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Follow the process of completing the square to solve 2x2 + 8x - 12 = 0.
Rom4ik [11]
Please:  Use "^" to denote exponentiation:  <span>2x^2 + 8x - 12 = 0

Reduce this by div. every term by 2:             </span><span>x^2 + 4x - 6 = 0

Here a=1, b=4 and c = -6.  Square half of b, obtaining (4/2)^2 = 4, and add, and then subtract, this 4 to x^2 + 4x - 6:

</span> x^2 + 4x +4  - 4 - 6 = 0.  Rewrite the square as (x+2)^2, obtaining new equation

(x+2)^2 = 10.  Take the sqrt of both sides:   x+2 = plus or minus sqrt(10).

Finally, solve for x:  x = -2 plus or minus sqrt(10).



8 0
3 years ago
The endpoints of are A(2, 2) and B(3, 8). is dilated by a scale factor of 3.5 with the origin as the center of dilation to give
goldfiish [28.3K]

Answer:

The given line segment whose end points are A(2,2) and B(3,8).

Distance AB is given by distance formula , which is

if we have to find distance between two points (a,b) and (p,q) is

=  \sqrt{(p-a)^2+(q-b)^2}

AB= \sqrt{(3-2)^2+(8-2)^2}=\sqrt{1+36}=\sqrt{37} = 6.08 (approx)

Line segment AB is dilated by a factor of 3.5 to get New line segment CD.

Coordinate of C = (3.5 ×2, 3.5×2)= (7,7)

Coordinate of D = (3.5×3, 3.5×8)=(10.5,28)

CD = AB × 3.5

CD = √37× 3.5

     = 6.08 × 3.5

= 21.28 unit(approx)

2. Slope of line joining two points (p,q) and (a,b) is given by

m=\frac{q-b}{p-a}

m= \frac{8-2}{3-2}=6

As the two lines are coincident , so their slopes are equal.

Slope of line AB=Slope of line CD = 6




6 0
3 years ago
Read 2 more answers
Frankie buys six and twelve fifteenths pounds of cherries and four and four sixths pounds of grapes. How many pounds of fruit di
Bas_tet [7]

Answer: I think it’s the 3rd one

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
In a right triangle, angle C measures 40. the hypotenuse of the triangle is 10 inches long. what is the approximate length of th
IrinaVladis [17]

Since angle C is not 90 degrees, it's one of the acute angles
in the right triangle.


               (The side adjacent to angle C ) divided by (hypotenuse)

is the cosine of angle C .       


                    (adjacent) / (10 inches)  =  cos(40 degrees)

Multiply each side by (10 inches):

                    Adjacent side = (10 inches) x cos(40) =

                                              (10 inches)  x  0.766  = 

                                                              7.66 inches   (rounded)
 
6 0
3 years ago
Read 2 more answers
Suppose that the sitting​ back-to-knee length for a group of adults has a normal distribution with a mean of mu equals 24.4 in.
solong [7]

Answer:

P(X \geq 26.6) = 0.0336, which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

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 problem, we have that:

\mu = 24.4, \sigma = 1.2

In this problem, a value x is significantly high if:

P(X \geq x) = 0.01

Using these​ criteria, is a​ back-to-knee length of 26.6 in. significantly​ high?

We have to find the probability of the length being 26.6 in or more, which is 1 subtracted by the pvalue of Z when X = 26.6. So

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

Z = \frac{26.6 - 24.4}{1.2}

Z = 1.83

Z = 1.83 has a pvalue of 0.9664.

1 - 0.9664 = 0.0336

P(X \geq 26.6) = 0.0336, which is greater than 0.01. So a back-to-knee length of 26.6 in. is not significantly high.

6 0
3 years ago
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