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2 years ago

Which expression is equivalent to 24 / ( 2 + x )?

Mathematics

1 answer:

elena-14-01-66 [18.8K]2 years ago

6 0

Answer:

Step-by-step explanation:

its the same but switched around

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Quadrilateral CAMP below is a rhombus. The length of PQ is (x + 2) units, and the length of QA is

Phoenix [80]

Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.

<u>Step-by-step explanation:</u>

As given by the statement in the problem,

Q may be the middle point, which cut the diagonal PA into 2 equal halves.

In rhombus, diagonals meet at right angles.

which means that PQ = QA

x+2 = 3x - 14

Grouping the terms, we will get,

3x -x = 14+ 2

2x = 16

dividing by 2 on both sides, we will get,

x = 16/2 = 8

8+2 = 3(8) - 14 = 10 = PQ or QA

3 0

3 years ago

Sophia‘s favorite homemade cookie recipe requires one cup of chocolate chips for 10 servings if the number of cups required for

weqwewe [10]

Answer:

3 cups

Step-by-step explanation:

We can use a proportion to find how many cups of chocolate chips she needs for 30 servings. Assuming c = cups of chocolate chips and b = batches

$\frac{c}{b}$

$\frac{1}{10} = \frac{c}{30}$

We can now multiply the diagonal values that don't include the missing variable (30 and 1) and then divide it by the value that is diagonal to the variable (10)

$30 \cdot 1 = 30\\30 \div 10 = 3$

Therefore, she needs 3 cups of chocolate chips to make 30 servings.

8 0

3 years ago

Read 2 more answers

HOW MANY SQUARE NUMBERS IN A PACK OF 52 CARDS

Vesnalui [34]

In a pack of 52 cards, there are 3 square numbers - 1, 4, and 9.

1 × 1 = 1
2 × 2 = 4
3 × 3 = 9

4 0

2 years ago

Evaluate the given integral by changing to polar coordinates. sin(x2 + y2) dA R , where R is the region in the first quadrant be

Leona [35]

In polar coordinates, the region $R$ is the set of points

$\left\{(r,\theta)\mid3\le r\le5,\,0\le\theta\le\dfrac\pi2\right\}$

and we have $x^2+y^2=r^2$ and $\mathrm dA=r\,\mathrm dr\,\mathrm d\theta$ . So the integral, converted to polar, is

$\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_0^{\pi/2}\int_3^5r\sin r^2\,\mathrm dr\,\mathrm d\theta=\frac\pi2\int_3^5r\sin r^2\,\mathrm dr$

Substitute $s=r^2$ to get

$=\displaystyle\frac\pi4\int_9^{25}\sin s\,\mathrm ds=\frac{(\cos9-\cos25)\pi}4$

5 0

3 years ago

If cos(θ) = 6/8 and θ is in the IV quadrant, then fine:

svetoff [14.1K]

Answer:

a) 1

b) $\frac{4}{3}$

c) = 1

Step-by-step explanation:

We are given the following in the question:

$\cos \theta = \dfrac{6}{8}$

θ is in the IV quadrant.

$\sin^2 \theta + \cos^2 \theta = 1\\\\\sin \theta = \sqrt{1-\dfrac{36}{64}} = -\dfrac{2\sqrt7}{8}\\\\\tan \theta = \dfrac{\sin \theta}{\cos \theta} = -\dfrac{2\sqrt7}{6}\\\\\csc \theta = \dfrac{1}{\sin \theta} = -\dfrac{8}{2\sqrt7}$

Evaluate the following:

$\tan \theta\times \cot \theta =\tan \theta\times\dfrac{1}{\tan \theta} = 1$

$\csc \theta\times \tan \theta\\\\= -\dfrac{8}{2\sqrt7}\times -\dfrac{2\sqrt7}{6} = \dfrac{4}{3}$

$\sin^2 \theta + \cos^2 \theta = 1\\\text{using the trignometric identity}$

5 0

3 years ago