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3 years ago

Need help now urgent because I am so confused on this problem and it’s due in 1 hour

Mathematics

2 answers:

adelina 88 [10]3 years ago

8 0

Answer:

Step-by-step explanation:

d would be correct because 5 1/2 is showing it is greater! (:

Nikolay [14]3 years ago

6 0

Answer:

We have less than 5 /12 cups of sugar

Step-by-step explanation:

c < 5 1/2

This means we have less than 5 1/2 of something

We have less than 5 /12 cups of sugar

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53.010,000 in scientific nottation

maks197457 [2]

I think it is 5.301 x 10^1

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3 years ago

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Workout the length of AB. Give your answers to 3 significant figures.

german

Answer:

AB ≈ 6.71

Step-by-step explanation:

Calculate the length using the distance formula

d = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }$

with (x₁, y₁ ) = A (- 1, 6) and (x₂, y₂ ) = B (5, 3)

AB = $\sqrt{(5-(-1))^2+(3-6)^2}$

= $\sqrt{(5+1)^2+(-3)^2}$

= $\sqrt{6^2+9}$

= $\sqrt{36+9}$

= $\sqrt{45}$

≈ 6.71 ( to 3 significant figures )

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3 years ago

Which of the following uses the distributive property correctly?

JulsSmile [24]

Answer:

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3( x + 5 ) = 3• x+ 3 • 5 uses the distributive property correctly.

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4 years ago

What is 14 divided by 25

Vladimir79 [104]

The answer to your question is 0.56

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4 years ago

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Special right triangles

garik1379 [7]

Answer: The answers are given below.

Step-by-step explanation: The calculations are as follows.

(1) We have in the given right-angled triangle,

$\tan 30^\circ=\dfrac{10}{x}\\\\\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{10}{x}\\\\\Rightarrow x=10\sqrt3,$

and

$y^2=(10)^2+x^2=100+(10\sqrt3)^2=100+300=400\\\\\Rightarrow y=20.$

∴ x = 10√3 and y = 20.

(2) We have in the given right-angled triangle,

$\cos 60^\circ=\dfrac{2}{x}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{2}{x}\\\\\Rightarrow x=4,$

and

$y^2=x^2-2^2=16-4=12\\\\\Rightarrow y=2\sqrt3.$

∴ x = 4 and y = 2√3.

(3) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{7}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{7}{y}\\\\\Rightarrow y=\dfrac{14\sqrt3}{3},$

and

$\sin 30^\circ=\dfrac{7}{x}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{7}{x}\\\\\Rightarrow x=14.$

∴ x = 14 and y = 14√3.

(4) We have in the given right-angled triangle,

$\sin 30^\circ=\dfrac{x}{6}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{6}\\\\\Rightarrow x=3,$

and

$\cos 30^\circ=\dfrac{y}{6}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{y}{6}\\\\\Rightarrow y=3\sqrt3.$

∴ x = 3 and y = 3√3.

(5) We have in the given right-angled triangle,

$\sin 30^\circ=\dfrac{y}{10}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{y}{10}\\\\\Rightarrow y=5,$

and

$\cos 30^\circ=\dfrac{x}{6}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{x}{10}\\\\\Rightarrow y=5\sqrt3.$

∴ x = 5 and y = 5√3.

(6) We have in the given right-angled triangle,

$\sin 30^\circ=\dfrac{x}{8}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{8}\\\\\Rightarrow x=4,$

and

$\cos 30^\circ=\dfrac{y}{8}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{y}{8}\\\\\Rightarrow y=4\sqrt3.$

∴ x = 4 and y = 4√3.

(7) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{7\sqrt3}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{7\sqrt3}{y}\\\\\Rightarrow y=14,$

and

$\sin 30^\circ=\dfrac{x}{y}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{14}\\\\\Rightarrow x=7.$

∴ x = 7 and y = 14.

(8) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{6\sqrt3}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{6\sqrt3}{y}\\\\\Rightarrow y=12$

and

$\sin 30^\circ=\dfrac{x}{y}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{12}\\\\\Rightarrow x=6$

∴ x = 6 and y = 12.

(9) We have in the given right-angled triangle,

$\cos 30^\circ=\dfrac{\sqrt3}{y}\\\\\Rightarrow \dfrac{\sqrt3}{2}=\dfrac{\sqrt3}{y}\\\\\Rightarrow y=2$

and

$\sin 30^\circ=\dfrac{x}{y}\\\\\Rightarrow \dfrac{1}{2}=\dfrac{x}{2}\\\\\Rightarrow x=4.$

∴ x = 4 and y = 2.

Thus, all are completed.

3 0

3 years ago