Ask question

Ask question

All categories

2 years ago

10

IfA=5 and B = 10, what is A* B equal to? O Two O 15 O 50 O Five

1 answer:

irina [24]2 years ago

7 0

Answer:

50

Step-by-step explanation:

5x10 = 50

You might be interested in

Enter the number in standard form.<br> 4.78 · 10-1

kogti [31]

The answer is 46.8
Plz mark me Brainly

8 0

2 years ago

Read 2 more answers

Please solve, thank you (and show steps):<br><br>Solve for x:<br><br>5x + 22 = 27

IrinaVladis [17]

Given the equation, 5x + 22 = 27:

The goal is to isolate the variable. Hence, a few mathematical operations are necessary to solve for the value of x.

Start by subtracting 22 both sides:
5x + 22 - 22 = 27 - 22
5x + 0 = 5
5x = 5

Next, divide both sides by 5 to isolate and solve for the value of x:

5x/5 = 5/5

x = 1

In order to verify whether x = 1 satisfies the give equation, we must substitute its value into the given equation:

5x + 22 = 27

5(1) + 22 = 27
5 + 22 = 27
27 = 27 (True statement. This implies that x = 1 is the correct value that satisfies the equation).

Therefore, the correct answer is x = 1.

6 0

2 years ago

Read 2 more answers

What is the equation of a circle with center (-2, 3) and radius 4?

Bad White [126]

Hello,

(x+2)²+(y-3)²=16

5 0

2 years ago

Can someone help me with this

aev [14]

Answer:

Its 0.90 per download.

Step-by-step explanation:

6 0

2 years ago

Using suitable algebraic identities, find the value of<br> 2004^2 - 2003^2 + 2002^2 - 2001^2

PSYCHO15rus [73]

$(x+1)^2-x^2=(x^2+2x+1)-x^2=2x+1$

Let $x=2003$ . Then

$2004^2-2003^2=(2003+1)^2-2003^2=2\cdot2003+1=4007$

Similarly, if $x=2001$ , then

$2002^2-2001^2=2\cdot2001+1=4003$

So

$2004^2-2003^2+2002^2-2001^2=4007+4003=8010$

Alternatively, you can consider the larger expression

$(x+3)^2-(x+2)^2+(x+1)^2-x^2$

Expanding each binomial product gives

$(x^2+6x+9)-(x^2+4x+4)+(x^2+2x+1)-x^2=4x+6$

Then if $x=2001$ , we get

$2004^2-2003^2+2002^2-2001^2=4\cdot2001+6=8010$

3 0

3 years ago