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2 years ago

12

Blueberries are $4.99 a pound. Jada buys p pounds of blueberries and pays c dollars.

1 answer:

Zepler [3.9K]2 years ago

5 0

Find out how much p pounds jada bought of blueberries and find out how much she paid

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Subtract 9x^2 - 7x from -4x^2 + 6

Rzqust [24]

Answer:

13x^2 - 7x + 6

Step-by-step explanation:

(9x^2 -7x) - 1(-4x^2 + 6)

(9x^2 -7x) + 4x^2 +6

9x^2 - 7x + 4x^2 + 6

(9x^2 + 4x^2) - 7x + 6

13x^2 - 7x + 6

8 0

3 years ago

What is the area of this parallelogram? A=31 ½Ft²<br> A = 35 ft²<br>A = 47½ ft²<br> A = 66 ½ ft²

Phoenix [80]

Answer:

A= 47 1/2

Step-by-step explanation:

3 0

1 year ago

A 51-foot wire running from the top of a tent pole to the ground makes an angle of 58° with the ground. If the length of the ten

AfilCa [17]

Answer:

<em>35.11 ft</em>

<em></em>

Step-by-step explanation:

This given situation can be thought of as triangle $\triangle PQR$ where PQ is the length of pole.

PR is the length of rope.

and QR is the distance of bottom of pole to the point of fastening of rope to the ground.

And $\angle Q \neq 90^\circ$

Given that:

PQ = 44 ft

PR = 51 ft

$\angle R = 58^\circ$

To find:

Side QR = ?

Solution:

We can apply Sine Rule here to find the unknown side.

Sine Rule:

$\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC}$

Where

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

$\dfrac{PR}{sinQ}=\dfrac{PQ}{sinR}\\\Rightarrow sin Q =\dfrac{PR}{PQ}\times sinR\\\Rightarrow sin Q =\dfrac{51}{44}\times sin58^\circ\\\Rightarrow \angle Q =79.41^\circ$

Now,

$\angle P +\angle Q +\angle R =180^\circ\\\Rightarrow \angle P +58^\circ+79.41^\circ=180^\circ\\\Rightarrow \angle P = 42.59^\circ$

Let us use the Sine rule again:

$\dfrac{QR}{sinP}=\dfrac{PQ}{sinR}\\\Rightarrow QR =\dfrac{sinP}{sinR}\times PQ\\\Rightarrow QR =\dfrac{sin42.59}{sin58}\times 44\\\Rightarrow QR = 35.11\ ft$

So, the answer is <em>35.11 ft</em>.

3 0

3 years ago

Quares and Differences of Squares

Otrada [13]

Answer:

hi

Step-by-step explanation:

hi

5 0

3 years ago

In the circle below, suppose m VUX = 152° and mZUVW = 77º. Find the following.

bezimeni [28]

Given:

In the circle, $m(\overarc{VUX})=152^\circ$ and $m(\angle MUV)=77^\circ$ .

To find:

The following measures:

(a) $m\angle VUX$

(b) $m\angle UXW$

Solution:

According to the central angle theorem, the central angle is always twice of the subtended angle intercepted on the same same arc.

$m(VUX)=2\times m\angle VWX$

$152^\circ=2\times m\angle VWX$

$\dfrac{152^\circ}{2}=m\angle VWX$

$76^\circ=m\angle VWX$

In a cyclic quadrilateral, the opposite angles are supplementary angles.

UVWX is a cyclic quadrilateral. So,

$m\angle VUX+m\angle VWX=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle VUX+76^\circ=180^\circ$

$m\angle VUX=180^\circ-76^\circ$

$m\angle VUX=104^\circ$

Now,

$m\angle UXW+m\angle UVW=180^\circ$ [Opposite angles of a cyclic quadrilateral]

$m\angle UXW+77^\circ =180^\circ$

$m\angle UXW=180^\circ-77^\circ$

$m\angle UXW=103^\circ$

Therefore, $m\angle VUX=104^\circ$ and $m\angle UXW=103^\circ$ .

7 0

2 years ago