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3 years ago

How do I start the equation?

Mathematics

2 answers:

algol [13]3 years ago

7 0

Answer:

x = 8

Step-by-step explanation:

40 + 6x + 2 = 90° because these angles are complementary

6x = 90 - 42

6x = 48 divide both sides by 6

x = 8

Mariana [72]3 years ago

4 0

Answer:

Step-by-step explanation:

Well they add up to a straight line which would be 180 degrees. That means 40, 6x+2 and 90 all add up to 180

40 + 90 + 6x + 2 = 180

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Shankar is teaching 2 courses, let us call them 435 and FYS. 435 has 3 sophomores, 8 juniors and 13 seniors; FYS has 5 sophomore

Mice21 [21]

Answer:

The probability that both students are of the same type is $\frac{149}{432}$ .

Step-by-step explanation:

The students in 435 are: {3 sophomores, 8 juniors and 13 seniors}

Number of students in 435 = 3 + 8 + 13 = 24

The students in FYS are: {5 sophomores, 7 juniors and 6 seniors}.

Number of students in FYS = 5 + 7 + 6 = 18

The teacher picks 1 student from each class.

The probability that both students are of the same type is:

P (Same type students) = P (Both are Sophomores) + P (Both are Juniors)

+ P (Both are Seniors)

= P (Sophomore ∩ Course 435) × P (Sophomore ∩ Course FYS)

+ P (Junior ∩ Course 435) × P (Junior ∩ Course FYS)

+ P (Senior ∩ Course 435) × P (Senior ∩ Course FYS)

$=[(\frac{3}{24} )\times(\frac{5}{18})]+[(\frac{8}{24} )\times(\frac{7}{18})]+[(\frac{13}{24} )\times(\frac{6}{18})]\\=\frac{15+56+78}{432}\\ =\frac{149}{432}$

Thus, the probability that both students are of the same type is $\frac{149}{432}$ .

3 0

3 years ago

The length of a rectangle is 3 1/6 cm to longer than its width If the perimeter of the rectangle is 15 1/3 cm, What is the width

alisha [4.7K]

The width and length of this rectangle are 3 7/12cm and 4 1/12cm respectively.

Hope this helped! ^v^

7 0

3 years ago

Graph triangle RST with vertices R(3, 7), S(-5, -2), and T(3, -5) and its image after a reflection over x = -3.

kirill115 [55]

Given:

The vertices of a triangle are R(3, 7), S(-5, -2), and T(3, -5).

To find:

The vertices of the triangle after a reflection over x = -3 and plot the triangle and its image on the graph.

Solution:

If a figure reflected across the line x=a, then

$(x,y)\to (-(x-a)+a,y)$

$(x,y)\to (-x+a+a,y)$

$(x,y)\to (2a-x,y)$

The triangle after a reflection over x = -3. So, the rule of reflection is

$(x,y)\to (2(-3)-x,y)$

$(x,y)\to (-6-x,y)$

The vertices of triangle after reflection are

$R(3,7)\to R'(-6-3,7)$

$R(3,7)\to R'(-9,7)$

Similarly,

$S(-5,-2)\to S'(-6-(-5),-2)$

$S(-5,-2)\to S'(-6+5,-2)$

$S(-5,-2)\to S'(-1,-2)$

And,

$T(3,-5)\to T'(-6-3,-5)$

$T(3,-5)\to T'(-9,-5)$

Therefore, the vertices of triangle after reflection over x=-3 are R'(-9,7), S'(-1,-2) and T'(-3,-5).

3 0

3 years ago

Which would be the next step? Does anyone know? Please, help.

Effectus [21]

Answer:

∠DAB = ∠DBA

Then AD=DB from above statement

7 0

3 years ago

Answer The equation 2x+4=8

allsm [11]

Answer:

x=2

Step-by-step explanation:

2x+4=8

2x+4-4=8-4

2x=4

2x/2=4/2

x=2

8 0

3 years ago

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