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2 years ago

Someone please help me

Mathematics

1 answer:

pentagon [3]2 years ago

8 0

Answer:

Option 3 or Option 4.

Step-by-step explanation:

I have this doubt because actually, a rhombus with a right angle (one) is not necessarily a square. Some rhombuses have one right angle and are still not squares. Now, if the four angles are right, it is a square, but this only says one.

On other hand, a parallelogram with congruent diagonals isn't a square necessarily either. Rectangles have congruent diagonals and are parallelograms.

In case you only can answer with one, I'll say is option 3, hope it helps!

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Find the circumference of a circle with radius, r = 6.8m. Give your answer rounded to 1 DP.

Assoli18 [71]

Answer:

42.7 m

Step-by-step explanation:

The Circumference of a circle is found by multiplying the radius of the circle by 2π. Here the radius is 6.8 m. The circumference of this circle is

C = 2π(6.8 m) = 13.6π m, or (to 1 decimal place, 42.7 m.

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3 years ago

Find -1 2/3 + 5/6 on the number line

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The answer would be be between 1 and 0.7

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3 years ago

Evaluate Dx / ^ 9-8x - x2^

Solnce55 [7]

It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for $\sqrt x$ .

In that case, you want to find the antiderivative,

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}$

Complete the square in the denominator:

$9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2$

Now substitute $x+4=5\sin y$ , so that $\mathrm dx=5\cos y\,\mathrm dy$ . Then

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy$

which simplifies to

$\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy$

Now, recall that $\sqrt{x^2}=|x|$ . But we want the substitution we made to be reversible, so that

$x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)$

which implies that $-\dfrac\pi2\le y\le\dfrac\pi2$ . (This is the range of the inverse sine function.)

Under these conditions, we have $\cos y\ge0$ , which lets us reduce $\sqrt{\cos^2y}=|\cos y|=\cos y$ . Finally,

$\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C$

and back-substituting to get this in terms of $x$ yields

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C$

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3 years ago

Rewrite 6 2/3 as an improper fraction

den301095 [7]

6 * 3 = 18 + 2 = 20

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3 years ago

Read 2 more answers

(this is separate)

leva [86]

Answer: 2/1

Step-by-step explanation:

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2 years ago