I need help I’m timed what is the y- intercept of the linear function? y= 1/2 x + 4

The sum of two numbers is 9.9, and the sum of the squares of the numbers is 53.21. What are the numbers?

Snezhnost [94]

Answer:

There are two possibilities:

$x_{1} = 3.5$ and $y_{1} = 6.4$

$x_{2} = 6.4$ and $y_{2} = 3.5$

Step-by-step explanation:

Mathematically speaking, the statement is equivalent to this 2-variable non-linear system:

$x + y = 9.9$

$x^{2} + y^{2} = 53.21$

First, $x$ is cleared in the first equation:

$x = 9.9 - y$

Now, the variable is substituted in the second one:

$(9.9-y)^{2} + y^{2} = 53.21$

And some algebra is done in order to simplify the expression:

$98.01-19.8\cdot y +2\cdot y^{2} = 53.21$

$2\cdot y^{2} -19.8\cdot y +44.8 = 0$

Roots are found by means of the General Equation for Second-Order Polynomials:

$y_{1} \approx \frac{32}{5}$ and $y_{2} \approx \frac{7}{2}$

There are two different values for $x$ :

$y = y_{1}$

$x_{1} = 9.9-6.4$

$x_{1} = 3.5$

$y = y_{2}$

$x_{2} = 9.9 - 3.5$

$x_{2} = 6.4$

There are two possibilities:

$x_{1} = 3.5$ and $y_{1} = 6.4$

$x_{2} = 6.4$ and $y_{2} = 3.5$

5 0

3 years ago

The value of x = 60° 5 x 30°

nikklg [1K]

Answer:

10

Step-by-step explanation:

with reference angle 30°

perpendicular (p) = 5

hypotenuse (h) = x

Now

sin 30° = p / h

1 / 2 = 5 / x

x = 10

Hope it will help :)

6 0

3 years ago

Quick guys, I need some help on this

worty [1.4K]

Answer:

I think c

(not sure though)

3 0

2 years ago

Find the point-slope equation for

mars1129 [50]

Answer:

y-24=m(x--4)

Step-by-step explanation:

point slope form:

y-y₁=m(x-x₁)

1) find the slope or "m" first:

slope formula: (y2-y1)/(x2-x1)

(-53-24)/7+4

-77/11=-7

slope=-7

so

y-y₁=-7(x-x₁)

y1=24

x1=-4

y-24=m(x--4)

hope this helps!

8 0

3 years ago

The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78

____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

$m-t_{a/2,n-1}\frac{s}{\sqrt{n} }$ ≤ μ ≤ $m+t_{a/2,n-1}\frac{s}{\sqrt{n} }$

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and $t_{a/2,n-1}$ is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and $t_{0.05,9}$ by 1.8331, we get that the 90% confidence interval for the mean speed is:

$74.2-(1.8331)\frac{5.3083}{\sqrt{10} }$ ≤ μ ≤ $74.2+(1.8331)\frac{5.3083}{\sqrt{10} }$

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0

2 years ago