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3 years ago

What is the y-intercept of an equation

Mathematics

2 answers:

Alona [7]3 years ago

8 0

Answer:

The y interpit means the y-axis. you should've known this in 5th grade.

Step-by-step explanation:

krek1111 [17]3 years ago

4 0

Answer:

The point on a graph where a line crosses the Y-axis (The vertical line)

Step-by-step explanation:

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for every 2 cars you wash your friend washes 3. you wash a total of 8 cars. how many does your friend wash?

daser333 [38]

Answer:

Step-by-step explanation:

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3 years ago

to convert a distance of 1.25 milesto feet,which ratio could you multiply by? remeber 1 mile=5290 feet

Alekssandra [29.7K]

What i got is 6,600. if its wrong, sorry bout that!

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4 years ago

Read 2 more answers

Find the area of each parallelogram. What is the relationship between the areas?

castortr0y [4]

Answer:

Area of parallelogram is given by:

$A = bh$

where b is the base and h is the height of parallelogram.

In parallelogram TQRS.

Coordinate of TQRS are;

T(8, 16), Q(4, 4), R(16, 4) and S(20, 16)

Coordinate of T'Q'R'S' are;

T'(2, 4), Q'(1, 1), R'(4, 1) and S'(5, 4)

Find the length of QR and PT:

Using distance(D) formula:

$D = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

$QR = \sqrt{(4-16)^2+(4-4)^2} =\sqrt{(-12)^2+0}= \sqrt{144}= 12$ units

Similarly;

For PT:

From the graph:

P(8, 4) and T(8, 16), then

$PT = \sqrt{(8-8)^2+(4-16)^2} =\sqrt{(-0)^2+(-12)^2}= \sqrt{144}= 12$ units

In parallelogram TQRS

PT represents the height and QR represents the base of the parallelogram respectively.

then;

Area of parallelogram TQRS = $QR \cdot PT$

⇒Area of parallelogram TQRS = $12 \cdot 12 = 144$ unit square.

Now, in parallelogram T'Q'R'S'

Q'R' represents the base and P'T' represents the height of the parallelogram respectively.

here, P'(2, 1)

Find the length of Q'R' and P'T':

$Q'R' = \sqrt{(1-4)^2+(1-1)^2} =\sqrt{(-3)^2+0}= \sqrt{9}= 3$ units

$P'T' = \sqrt{(2-2)^2+(1-4)^2} =\sqrt{(-0)^2+(-3)^2}= \sqrt{9}= 3$ units

Then;

Area of parallelogram T'Q'R'S' = $Q'R' \cdot P'T'$

Area of parallelogram T'Q'R'S' = $9 \cdot 9= 81$ unit square.

Now, we have to find the relationship between the areas.

$\frac{\text{Area of parallelogram TQRS}}{\text{Area of parallelogram T'Q'R'S'}} = \frac{144}{81}$

then;

the relationship between the areas of TQRS and T'Q'R'S' is:

$\text{Area of parallelogram TQRS} = \frac{144}{81} \cdot {\text{Area of parallelogram T'Q'R'S'}$

7 0

3 years ago

46. A holiday resort in Tasmania reduced its overnight rates from 233 to 159.

Debora [2.8K]

The link takes you to it

3 0

3 years ago

Find the degree measure of each angle in the triangle.

Alex777 [14]

Answer:

∠R = 56°
∠Q = 90°
∠S = 34°

Step-by-step explanation:

The given triangle is a right angled triangle.

So, the angles in the triangle are :

90°
(2x + 38)°
(5x - 11)°

Solving according to <u>angle sum property</u>,

Sum of all angles in a triangle is 180°

$\longrightarrow$ 90° + (2x + 38)° + (5x - 11)° = 180°

$\longrightarrow$ 117° + 7x = 180°

$\longrightarrow$ 7x = 180° - 117°

$\longrightarrow$ 7x = 63°

$\longrightarrow$ x = 9

Angles =

$\longrightarrow$ 2(9) + 38

$\longrightarrow$ 56°

$\longrightarrow$ 5(9) - 11

$\longrightarrow$ 34°

∠R = 56°
∠Q = 90°
∠S = 34°

The angles are 56°, 90° and 34°.

4 0

3 years ago