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3 years ago

Can someone help? Please

Mathematics

2 answers:

Flauer [41]3 years ago

8 0

Answer:

x=1

Step-by-step explanation:

3(x)+1 substitute 1 in for x

3(1)+1=4

10(x)-6 substitute 1 in for x

10(1)-6=4

4=4

(sorry if that doesn't make any sense)

Serjik [45]3 years ago

7 0

Answer:

x=1

Step-by-step explanation:

3x+1=10x-6

3x+7=10x

7=7x

1=x

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If the areas of the given pairs of shapes are equal, find the value of x.

den301095 [7]

Answer:

$x = 8cm$

Step-by-step explanation:

we are given a square and rectangle.we want to figure out x which is the width of the rectangle. we are also given a condition i.e

the area of the square equal to the area of the rectangle

therefore,

$\displaystyle \rm A _{square } = A _{rect}$

recall the formula of the area of square and rectangle so,

$\displaystyle {s}^{2} = lw$

now assign variables

$s \implies16cm$
$l \implies32cm$
$w \implies x$

thus substitute:

$\displaystyle 32cmx = {16cm}^{2}$

simplify square:

$\displaystyle 32cmx = 256cm^2$

divide both sides by 32:

$\displaystyle \boxed{x = 8cm}$

and we're done!

3 0

2 years ago

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Two teachers are taking 8 students to the aquarium. Adult admission is $12, but students receive a discount of $3 off

zhuklara [117]

Answer:

86$

12-3=9

9x8=72

72+14=86$

6 0

3 years ago

Really easy question !! :)

kherson [118]

Answer:

20:21

Step-by-step explanation:

Let's first start by converting each ratio so that the number of girls is the same

2:3

4:7

The least common multiple of 3 and 4 is 12 so...

8:12

12:21

We add 8 and 12 to get the number of junior members...

8+12=20

Then we put it in a ratio with the amount of adult members...

20:21

5 0

3 years ago

Can someone help me, please!!

love history [14]

Id say its option 3.

5 0

2 years ago

What are the possible values of x in x2 + 3x + 3 = 0?

e-lub [12.9K]

To solve a quadratic equation like $ax^2+bx+c=0$ , you can use the quadratic formula

$x_{1,2} = \cfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

In your case, $a = 1, b = c = 3$ , so the formula becomes

$x_{1,2} = \cfrac{-3\pm\sqrt{(-3)^2-4\cdot 1\cdot 3}}{2\cdot 1}$

We can simplify the expression:

$x_{1,2} = \cfrac{-3\pm\sqrt{9-12}}{2} = \cfrac{-3\pm\sqrt{-3}}{2}$

Since -3 is negative, its square root is computed as

$\sqrt{-3} = \sqrt{-1\cdot 3} = \sqrt{-1}\sqrt{3} = \sqrt{3}i$

So, the solutions are

$x = \cfrac{-3+i\sqrt{3}}{2} \text{ or } x = \cfrac{-3-i\sqrt{3}}{2}$

4 0

3 years ago

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