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3 years ago

Write an equation that represents the line passing through the point -3, -1 with a slope of 4

Mathematics

1 answer:

Alex17521 [72]3 years ago

4 0

Answer:

Equation that represents the line passing through the point -3, -1 with a slope of 4 is

$y= 4x+11$

Step-by-step explanation:

Equation of line in point slope form is

$y-y_1=m(x-x_1)$ -------(1)

Here

$m=4 \ \ \ and \ \ \ (x_1, y_1) = (-3, -1)$

Substituting values in equation (1)

$y+1=4(x+3)\\ y= 4x+11$

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Find BC please in the picture

tatiyna

B. 25 Km. The measure of BC is 25 km.

The easiest way to solve this problem is using the cosine theorem c = √a²+b²-2ab*cos A.

BC = √AC²+AB²-2(AC)(AB)*cos A

BC = √(21km)²+(14km)²-2(21km)(14km)*cos 89°

BC = √441km²+196km²-588km²*(0.017)

BC =√637km²-10.26km²

BC = √636.74km²

BC = 25.03km ≅ 25

7 0

3 years ago

Plzzz help plzzzzzzzzzzzzzz

mezya [45]

Given:

The figure of two quadrilaterals.

In $ABCD,AB=18,BC=20,CD=22,AD=24$

In $EFGH,EF=27,FG=30,GH=34, EH=36$

To find:

Whether the figures are congruent, similar or neither.

Solution:

Ratio of corresponding sides are:

$\dfrac{AB}{EF}=\dfrac{18}{27}$

$\dfrac{AB}{EF}=\dfrac{2}{3}$

Similarly,

$\dfrac{BC}{FG}=\dfrac{20}{30}$

$\dfrac{BC}{FG}=\dfrac{2}{3}$

$\dfrac{CD}{GH}=\dfrac{22}{34}$

$\dfrac{CD}{GH}=\dfrac{11}{17}$

And,

$\dfrac{AD}{EH}=\dfrac{24}{36}$

$\dfrac{AD}{EH}=\dfrac{2}{3}$

Clearly, $\dfrac{AB}{EF}=\dfrac{BC}{FG}=\dfrac{AD}{EH}\neq \dfrac{CD}{GH}$ .

All corresponding sides are not proportional.

Therefore, the figures are neither similar nor congruent. Hence, third option is correct.

7 0

3 years ago

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

yKpoI14uk [10]

Answer:

a) 0.5588

b) 0.9984

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 24

Standard Deviation, σ = 6.4

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(score between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{6.4} \leq z \leq \displaystyle\frac{30-24}{6.4}) = P(-0.62 \leq z \leq 0.94)\\\\= P(z \leq 0.94) - P(z < -0.62)\\= 0.8264 - 0.2676 = 0.5588 = 55.88\%$

$P(20 \leq x \leq 30) = 55.88\%$

b) Sampling distribution

Sample size, n = 22

The sample will follow a normal distribution with mean 24 and standard deviation,

$s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{6.4}{\sqrt{22}} =1.36$

c) P(mean score of sample is between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{1.36} \leq z \leq \displaystyle\frac{30-24}{1.36}) = P(-2.94 \leq z \leq 4.41)\\\\= P(z \leq 4.41) - P(z < -2.94)\\= 1 - 0.0016 = 0.9984 = 99.84\%$