All categories

2 years ago

Find the area of parallelogram RSTU.

Mathematics

2 answers:

uysha [10]2 years ago

6 0

Answer:

252.5

Step-by-step explanation:

{Be careful}

The area: $15.3\times16.5\times\frac{1}{2}\times2$

$=252.5$ inches²

I hope this helps you

:)

GuDViN [60]2 years ago

4 0

$\huge\sf\blue{Solution:-}$

Here is a Parallelogram RSTU.

Given,

Height = 16.5 inches

Base = 15.3 inches

★Area of Parallelogram is = Base × Height

So, Area = 15.3 × 16.5 => 252.45 inches² or 252.5 inches²

________________________________________

You might be interested in

Which choice names a diameter of circle f

nadezda [96]

It would be c because fe is a radius ba is a chord and ec isnt a line

5 0

3 years ago

Give the rule for the translation if the coordinates of the preimage are A(2,-4) B(5,-4) C(5,-2) D(4,-2) E(4,-1) F(3,-1) G(3,-2)

trasher [3.6K]

Answer:

(x, y) ⇒ (x -7, y)

Step-by-step explanation:

The y-coordinates are unchanged. Each x-coordinate of the image is 7 units less than the corresponding pre-image coordinate. As a rule, this is ...

(x, y) ⇒ (x -7, y)

6 0

2 years ago

Read 2 more answers

a submarine dives below the surface, heading downward in three moves.if each move downward was 325 feet, where is the submarine

Degger [83]

If one move is 325 and there's 3 moves you have to multiply them together to get 975 but since the submarine is going down its a negative integer so the answer would be -975

5 0

3 years ago

Read 2 more answers

A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set

lara [203]

A circle with center $(3,0)$ and radius $5$ has equation

$(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x = 16$

If we substitute $y=2x+k$ in this equation, we have

$x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0$

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

$\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0$

The expression simplifies to

$-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89$

The solutions to the associated equation are

$k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}$

So, the parabola is negative between the two solutions:

$-6-5\sqrt{5}$

8 0

3 years ago

Can you give me different ways to solve 10 divided by 4

maxonik [38]

Write it as a fraction then simplify. Use a calculator.

4 0

2 years ago