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2 years ago

Can someone help me !! And PLEASE SHOW YOUR WORK

Mathematics

1 answer:

Alona [7]2 years ago

7 0

The formula for the area of a trapezoid is A=1/2(base one+base two)h
All you have to do is find the base one, base two, and height using the given information and then plug it into the equation.

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AWARDING DOUBLE POINTS question in pic, will report anything other than the answer

mash [69]

Answer:

m∠R is 72°

Step-by-step explanation:

In the given figure

∵ ΔPQR ≅ ΔUVW

→ From congruency

∵ m∠P = m∠U

∵ m∠Q = m∠V

∴ m∠R = m∠W

∵ m∠R = (10x - 18)°

∵ m∠W = 8x°

∵ m∠R = m∠W

→ Equate their measures

∴ 10x - 18 = 8x

→ Add 18 to both sides

∵ 10x - 18 + 18 = 8x + 18

∴ 10x = 8x + 18

→ Subtract 8x from both sides

∴ 10x - 8x = 8x - 8x + 18

∴ 2x = 18

→ Divide both sides by 2 to find x

∴ x = 9

→ Substitute the value of x in the m∠R

∵ m∠R = 10(9) - 18

∴ m∠R = 90 - 18

∴ m∠R = 72°

∴ m∠R is 72°

7 0

2 years ago

I don't know how to do these sort of problems... Can someone help me out?

VLD [36.1K]

180=76+4x
180-76=4x
104=4x
104/4=x
26=x
Measurement A= 26

6 0

2 years ago

QUESTION 4 PLEASE HELP ME ASAP

aniked [119]

Answer:

Part 1: The entry fee is $5

Part 2: Each minute costs 25 cents

Step-by-step explanation:

In the expression, the '+5' represents the fee added to '.25m' the cost per each minute.

4 0

1 year ago

Grocery store sella kiwis for $4.25 a pound and Asian pears for $2.25 a pound. Write an equation in standard form using integers

bogdanovich [222]

The pears could equal exactly $18
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4 0

3 years ago

F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa

DerKrebs [107]

a. Parameterize $C$ by

$\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath$

with $0\le t\le1$ .

b/c. The line integral of $\vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath$ over $C$ is

$\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt$

$=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt$

$=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt$

$=\displaystyle10\int_0^1\mathrm dt=\boxed{10}$

d. Notice that we can write the line integral as

$\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)$

By Green's theorem, the line integral is equivalent to

$\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy$

where $D$ is the triangle bounded by $C$ , and this integral is simply twice the area of $D$ . $D$ is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0

2 years ago