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2 years ago

Simplify b^10/b^2 OA. b^-8 OB. b^5 O c. b^-5 O D. b^8

Mathematics

1 answer:

Allushta [10]2 years ago

7 0

Step-by-step explanation:

The answer is D)

b^10/b^2

= b^10-b^2 = b^8

hope this helps.pls like and follow if you understand.

thanks:-):-)

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Which expression is the right expression?

Juli2301 [7.4K]

D, you have to factor 3 out of 6x squared

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3 years ago

Which value, when placed in the box, would result in a system of equations with infinitely many solutions?

Fantom [35]

Answer:

<h2>If we placed the number 10 in the box, we obtain a system of equations with infinitely many solutions.</h2>

Step-by-step explanation:

The given system is

$y=2x-5\\2y-4x=a$

<em>It's important to know that a system with infinitely many solutions, it's a system that has the same equation</em>, that is, both equation represent the same line, or as some textbooks say, one line is on the other one, so they have inifinitely common solutions.

Having said that, the first thing we should do here is reorder the system

$y=2x-5\\2y=4x+a$

This way, you can compare better both equations. If you look closer, observe that the second equation is double, that is, it can be obtained by multiplying a factor of 2 to the first one, that is

$y=2x-5\\2y=4x-10$

So, by multiplying such factor, we obtaine the second equation. Observe that $a$ must be equal to 10, that way the system would have infinitely solutions.

Therefore, the answer is 10.

4 0

3 years ago

Help me please i cant get it

Leni [432]

Answer:

Let's define the cost of the cheaper game as X, and the cost of the pricer game as Y.

The total cost of both games is:

X + Y

We know that both games cost just above AED 80

Then:

X + Y > AED 80

From this, we want to prove that at least one of the games costed more than AED 40.

Now let's play with the possible prices of X, there are two possible cases:

X is larger than AED 40

X is equal to or smaller than AED 40.

If X is more than AED 40, then we have a game that costed more than AED 40.

If X is less than or equal to AED 40, then:

X ≥ AED 40

Now let's take the maximum value of X in this scenario, this is:

X = AED 40

Replacing this in the first inequality, we get:

X + Y > AED 80

Replacing the value of X we get:

AED 40 + Y > AED 80

Y > AED 80 - AED 40

Y > AED 40

So when X is equal or smaller than AED 40, the value of Y is larger than AED 40.

So we proven that in all the possible cases, at least one of the two games costs more than AED 40.

7 0

3 years ago

0-6

dybincka [34]

<h3>Given :-</h3>

Pranav ran 20.3 km more than Branda

Pravin ran 38.6 km

$\\ \\$

To find :

No. of kilometers did Brenda run.

$\\ \\$

No. of kilometers ran by Brenda be x.

$\\ \\$

<h3>Solution:</h3>

$\\ \\$

Equation formed:-

$\\ \\$

Total distance covered by Pravan = More distance covered by Pravan + distance covered by Brenda.

Therefore:-

$\\ \\$

$\leadsto \sf38.6 = 20.3 + x$

Write the equation

$\\ \\$

$\leadsto \sf38.6 - 20.3 =x$

When we transfer 20.3 to left side the positive sign (+) will change into negative sign (–)

$\\ \\$

$\leadsto \sf x = 38.6 - 20.3$

Arrange the equation because x is always represented at left side.

$\\ \\$

$\leadsto \boxed {\pmb{\sf x = 18.3}}\star$

After subtracting 38.6 with 20.3 we will get result as 18.3 .

$\\ \\$

$\therefore \red{ \underline{ \pmb{\frak{Distance ~ covered ~by ~Brenda~ is ~equal ~to~18.3~kilometers}}}}$

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2 years ago

AC−⇀− bisects ∠BAD.If m∠BAC=38°, find the m∠BAD.

PilotLPTM [1.2K]

Answer:

180-38=142 the other angle is 142

Step-by-step explanation:

4 0

3 years ago