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3 years ago

Put this equation in slope intercept form: 3x - 2y = -6

Mathematics

1 answer:

VashaNatasha [74]3 years ago

8 0

Answer:

Y= 3/2x+3

Step-by-step explanation:

Hopefully this helps if so please mark as brainliest

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What is the equation

erastova [34]

Thats figure is made of 2 triangles and 1 rectangle
so we will find the area of triangles and rectangles then add them

Triangle Area=1/2*x/2*2=x/2

multiply by 2 as we have two triangles=2*x/2=x
Area of a rectangle=x*2=2x
Add them=x+2x=14(given area of trepaziod
3x=14

hope i am correct

6 0

3 years ago

In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.

bezimeni [28]

Given:

In the circle, $m(\overarc{VUX})=152^\circ$ and $m(\angle MUV)=77^\circ$ .

To find:

The following measures:

(a) $m\angle VUX$

(b) $m\angle UXW$

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

$m(VUX)=2\times m\angle VWX$

$152^\circ=2\times m\angle VWX$

$\dfrac{152^\circ}{2}=m\angle VWX$

$76^\circ=m\angle VWX$

In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

$m\angle VUX+m\angle VWX=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle VUX+76^\circ=180^\circ$

$m\angle VUX=180^\circ-76^\circ$

$m\angle VUX=104^\circ$

Now,

$m\angle UXW+m\angle UVW=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle UXW+77^\circ =180^\circ$

$m\angle UXW=180^\circ-77^\circ$

$m\angle UXW=103^\circ$

Therefore, $m\angle VUX=104^\circ$ and $m\angle UXW=103^\circ$ .

7 0

2 years ago

Find all real zeros of the function f(x)=-5x(x^2-49)(x-2)

german

The zeros are 0,7,-7,2

4 0

3 years ago

Please help me please will give brainliest to

uranmaximum [27]

Answer:

<h3>step by step explanation</h3>

I hope it's help

7 0

2 years ago

From a large number of actuarial exam scores, a random sample of scores is selected, and it is found that of these are passing s

Mnenie [13.5K]

<u>Supposing 60 out of 100 scores are passing scores</u>, the 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $\frac{1+\alpha}{2}$ .

60 out of 100 scores are passing scores, hence $n = 100, \pi = \frac{60}{100} = 0.6$

95% confidence level

So $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so $z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 - 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.5$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6 + 1.96\sqrt{\frac{0.6(0.4)}{100}} = 0.7$

The 95% confidence interval for the proportion of all scores that are passing is (0.5, 0.7).

The lower limit is 0.5.
The upper limit is 0.7.

A similar problem is given at brainly.com/question/16807970

5 0

2 years ago