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2 years ago

NEED HELP ASAP WORTH LOTS OF POINTS

Mathematics

2 answers:

Vikentia [17]2 years ago

7 0

Statement:

PQRS is a rhombus with PQ = (9b - 5) and QR = (3b - 0.8) meters. S is the midpoint of RT.

To find out:

The length of RT.

Solution:

There is a property of a rhombus that all sides of the rhombus are equal.

So, PQ = QR = RS.

Therefore, 9b - 5 = 3b - 0.8
or, 9b - 3b = -0.8 + 5
or, 6b = 4.2
or, b = 4.2 ÷ 6 = 0.7 m

Therefore, the length of RS
= 9b - 5
= 9(0.7) - 5 m
= 6.3 - 5 m
= 1.3 m

S is the midpoint of RT.
Hence, TS = RS = 1.3 m

Therefore, the measure of RT
= TS + RS
= (1.3 + 1.3) m
= 2.6 m

Answer:

RT = 2.6 meters

Hope you could understand.

If you have any query, feel free to ask.

otez555 [7]2 years ago

6 0

Answer: 2.6

Step-by-step explanation:

PQ = QR (because it’s a rhombus), so

9b - 5 = 3b - 0.8

b = 0.7

And RS = PQ (again because rhombus), so RS = 9b - 5 = 1.3. Since RT is twice RS, we have

RT = 2(1.3) = 2.6.

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Leviafan [203]

Answer:

The equation determine a relation between x and y

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The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* Lets solve the problem

- The equation of a circle of center (h , k) and radius r is

(x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒ (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

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in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

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3 years ago

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3 years ago

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3 years ago

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3 years ago

At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

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Let's begin by listing out the information given to us:

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$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

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$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

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