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3 years ago

Find angle E if angle E and angle F are supplementary, angle E= 2x + 5 and angle F= 5x - 38

Mathematics

1 answer:

Bingel [31]3 years ago

4 0

Answer:

About 65.86 degrees

Step-by-step explanation:

A supplementary angle means that the angles add to 180 degrees.

Angle E + Angle F = 180 degrees.

You can then plug in the given equations for E and F.

2x + 5 + 5x - 38 = 180

Combine like terms:

2x + 5x = 7x

5 - 38 = -33

New equation: 7x-33=180

Remove the -33 from the left side of the equation by adding 33 to each side

7x-33+33 = 180+33 ---> 7x = 213

Divide by 7 on each side

7x/7 = 213/7 ---> x = about 30.43

You can then plug this into your equation for E and solve

2(30.43) + 5 = 65.86 Degrees

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Brady made a scale drawing of a rectangular swimming pool on a coordinate grid. The points (-20, 25), (30, 25), (30, -10) and (-

djverab [1.8K]

Answer:

Length = 50 units

width = 35 units

Step-by-step explanation:

Let A, B, C and D be the corner of the pools.

Given:

The points of the corners are.

$A(x_{1}, y_{1}})=(-20, 25)$

$B(x_{2}, y_{2}})=(30, 25)$

$C(x_{3}, y_{3}})=(30, -10)$

$D(x_{4}, y_{4}})=(-20, -10)$

We need to find the dimension of the pools.

Solution:

Using distance formula of the two points.

$d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ ----------(1)

For point AB

Substitute points A(30, 25) and B(30, 25) in above equation.

$AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}$

$AB=\sqrt{(30+20)^{2}}$

$AB=\sqrt{(50)^{2}$

AB = 50 units

Similarly for point BC

Substitute points B(-20, 25) and C(30, -10) in equation 1.

$d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}$

$BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}$

$BC=\sqrt{(-35)^{2}}$

BC = 35 units

Similarly for point DC

Substitute points D(-20, -10) and C(30, -10) in equation 1.

$d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}$

$DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}$

$DC=\sqrt{(30+20)^{2}}$

$DC=\sqrt{(50)^{2}}$

DC = 50 units

Similarly for segment AD

Substitute points A(-20, 25) and D(-20, -10) in equation 1.

$d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}$

$AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}$

$AD=\sqrt{(-20+20)^{2}+(-35)^{2}}$

$AD=\sqrt{(-35)^{2}}$

AD = 35 units

Therefore, the dimension of the rectangular swimming pool are.

Length = 50 units

width = 35 units

7 0

3 years ago

Solve using the quadratic formula 2x^2 + 7x - 15 = 0

marissa [1.9K]

Answer:

1.5 or -5

Step-by-step explanation:

2x²+7x-15=0

Divide by 2:

x²+7x/2-15/2=0

x=-7/2/2 ± √49/4(4)+15/2

=-7/4 ± √49/16+120/16

=-7/4± √169/16=-7/4 ±13/4

=6/4 or -20/4

=1.5 or -5

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3 years ago

Select the postulate that is illustrated for the real numbers. 2(x + 3) = 2x + 6 The commutative postulate for multiplication Mu

madam [21]

Answer:

Step-by-step explanation:

Postulate 5: If two planes intersect, then their intersection is a line.

4 0

3 years ago

Read 2 more answers

You put $200 in a savings account. The account earns 2% simple interest

Black_prince [1.1K]

Answer:

interest total after 5 years is $20.82

Step-by-step explanation:

6 0

3 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

7 0

2 years ago