All categories

3 years ago

PLS HELP!! I WILLGIVE BRAINLIEST

Mathematics

2 answers:

Alenkinab [10]3 years ago

6 0

58 hope it will help you

rjkz [21]3 years ago

5 0

The answer is -58 as it’s gone from 54 to -4

You might be interested in

A school is painting its logo in the shape of a triangle in the middle of its sports field. The school wants the height of a tri

vova2212 [387]

Answer:

The base of the logo must be less than 6 feets but greter than 0 feet, 0<b<6.

Step-by-step explanation:

The formula of area of triangle is A = 1/2×b×h where b is the base and h is the height. It is given that the height of the logo is 4ft and the area is lesa than 12 ft². Using this information, we can make an inequality exrpression :

Let h = 4ft

1/2 × b × 4 < 12

1/2 × b < 12 ÷ 4

1/2 × b < 3

b < 3 ÷ 1/2

b < 6 ft

3 0

3 years ago

What is the other 0 of the quadratic function

adelina 88 [10]

Answer: Unknown

Step-by-step explanation:

-b+/-√b^2-4ac/2a < plug a b and c into this equation and boom done ez

7 0

2 years ago

Solve for c. Solve for x. Solve for c. Solve for x. Solve for x.

Mariana [72]

Answer:

1) The option $c=9-a+b$ is correct.

2) The option x=-2 or x=-3 is correct.

3) The option $c=9-a+b$ is correct.

4) The option $x=\frac{11}{3}$ is correct.

5) The option x=-2 or x=-3 is correct.

Step-by-step explanation:

1) Given equation is $\sqrt{a-b+c}=3$

Now to solve the equation for c:

$\sqrt{a-b+c}=3$

Squaring on both sides we get

$(\sqrt{a-b+c})^2=3^2$

$a-b+c=9$

$c=9-a+b$

Therefore the option $c=9-a+b$ is correct.

2) Given equation is $\sqrt{3x+10}=x+4$

Now to solve the equation for x:

$\sqrt{3x+10}=x+4$

Squaring on both sides we get

$(\sqrt{3x+10})^2=(x+4)^2$

$3x+10=x^2+8x+16$

$x^2+8x+16-3x-10=0$

$x^2+5x+6=0$ which is a quadratic equation in x.

We can solve it by finding factors

$x^2+5x+6=(x+2)(x+3)$

$(x+2)(x+3)=0$

x+2=0 or x+3=0

Therefore x=-2 or x=-3

Therefore the option x=-2 or x=-3 is correct.

3) Given equation is $\sqrt{a-b+c}=3$

Now to solve the equation for c:

$\sqrt{a-b+c}=3$

Squaring on both sides we get

$(\sqrt{a-b+c})^2=3^2$

$a-b+c=9$

$c=9-a+b$

Therefore the option $c=9-a+b$ is correct.

4) Given equation is $\sqrt{x+3}=2\sqrt{x-2}$

Now to solve the equation for x:

$\sqrt{x+3}=2\sqrt{x-2}$

Squaring on both sides we get

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

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

$x+3=4(x-2)$

$x+3=4x-8$

$4x-8-x-3=0$

$3x-11=0$

$x=\frac{11}{3}$

Therefore the option $x=\frac{11}{3}$ is correct.

5) Given equation is $\sqrt{3x+10}=x+4$

Now to solve the equation for x:

$\sqrt{3x+10}=x+4$

Squaring on both sides we get

$(\sqrt{3x+10})^2=(x+4)^2$

$3x+10=x^2+8x+16$

$x^2+8x+16-3x-10=0$

$x^2+5x+6=0$ which is a quadratic equation in x.

We can solve it by finding factors

$x^2+5x+6=(x+2)(x+3)$

$(x+2)(x+3)=0$

x+2=0 or x+3=0

Therefore x=-2 or x=-3

Therefore the option x=-2 or x=-3 is correct.

5 0

3 years ago

Read 2 more answers

Need help. i learned this already i jus need a refresher. thanks

Arisa [49]

Answer:

Step-by-step explanation:

4/5

4 0

3 years ago

How do you subtract this? 91147-38350

Orlov [11]

The answer is 52797. just line everything up vertically and subtract left to right, make sure you carry your remainders if any

6 0

3 years ago