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3 years ago

Divide. 15÷−5 Drag and drop the correct number into the box to complete the sentence. The quotient is Response area.

Mathematics

1 answer:

german3 years ago

8 0

Answer:

15÷-5=-3

Step-by-step explanation:

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I NEED HELP!!!!!! 4x+2x-5=7x-1

yan [13]

Let's solve your equation step-by-step.

4x+2x−5=7x−1

Step 1: Simplify both sides of the equation.

4x+2x−5=7x−1

4x+2x+−5=7x+−1

(4x+2x)+(−5)=7x−1(Combine Like Terms)

6x+−5=7x−1

6x−5=7x−1

Step 2: Subtract 7x from both sides.

6x−5−7x=7x−1−7x

−x−5=−1

Step 3: Add 5 to both sides.

−x−5+5=−1+5

−x=4

Step 4: Divide both sides by -1.

−x

−1

x=−4

Answer:

x=−4

i hope this helps you

3 0

3 years ago

Please Help!!!!

Zinaida [17]

Answer:

1 no association

Number 2 non inear

3 0

3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0

3 years ago

Find the area and circumference of the 2 circles

velikii [3]

Answer:

32, 45

Step-by-step explanation:

4 0

3 years ago

(6×1,000)+(1×100)+(2×10)+(9× 10 1 )+(4× 100 1 )=?

stiv31 [10]

Answer:

11033

Step-by-step explanation:

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2 years ago