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2 years ago

Solve the system of equations 10×+7y=99andy-×=2?

1 answer:

8 0

You can use substitution and solve y-x=2 for y which is y= x+ 2 and plug it into the y of the other equasion and you get x=5 y=7

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Suppose y varies directly with x. If y = –20 when x = 2, find y when x = 14

Natasha_Volkova [10]

Answer:

y = -140

Step-by-step explanation:

Y varies directly with x

y = kx

We know y = -20 when x =2, we can solve for k

-20 = k*2

Divide each side by 2

-20/2= 2k/2

-10 =k

y = -10x

Now we want to find y when x = 14

y = -10 *14

y = -140

3 0

3 years ago

Read 2 more answers

Which Expression is equivalent to 4(6 + x + 3x)

Gennadij [26K]

Answer:

=4(6) + 4(X) + 4(3x)

=24+4x+12x

=24+16x

7 0

2 years ago

I need help on this question, not to good with graph. Anybody help pleaseee!!

LenKa [72]

y = -3x - 2

Since a < 0 we have a decreasing line, so it can be only the first or the last

And if we solve the equation for y = 0 we have

0 = -3x - 2

3x = -2

x = -2/3

So, the last one is the right

3 0

2 years ago

Solve: 2a-3=9 Solve: -3(b+2)= -9

Softa [21]

The Solution are : a=3.
b=1.

3 0

2 years ago

Read 2 more answers

Both the La Plata river dolphin (Pontoporia blainvillei) and

Citrus2011 [14]

Answer:

1. A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin; 2. The radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball; 3. The volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.

Step-by-step explanation:

If we expressed a number as:

$\\ N = a * 10^{b}$ (1)

Where

$\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10}$ (2)

$\\ 1 \leq a < 10$ (3)

Then, b represents the order of magnitude of such a number (Order of magnitude (2020), in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The La Plata river dolphin weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the scientific notation):

$\\ 30kg \leq Dolphin_{weight} \leq 50kg$

$\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg$

$\\ 35000kg \leq Whale_{weight} \leq 40000kg$

$\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg$

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

$\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)$

$\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)$

Then

$\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}$

Thus

A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin.

<h3>Second case</h3>

Following the same reasoning, we can conclude that the radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball:

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}$