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2 years ago

Please help if Yk the answer.

Mathematics

2 answers:

yKpoI14uk [10]2 years ago

4 0

Answer:25

Step-by-step explanation:

lora16 [44]2 years ago

3 0

Answer - B

60 + 4x + 20 = 180

4x = 100

x = 25

180 - 60 = 120

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Here are two squares, A and B. A B The length of each side of square B is 4cm greater than the length of each side of square A.

alina1380 [7]

Answer:

144cm²

Step-by-step explanation:

Area of square A Aa = La²

Area of square B Ab= Lb²

If the length of each side of square B is 4cm greater than the length of each side of square A, then;

Lb = La + 4 ... 1

Also if the area of square B is 70cm² greater than the area of square A, then;

Ab = Aa + 70

Since Aa = La² and Ab = Lb²

Lb² = La² + 70 ...2

Substitute 1 into 2;

(La+4)² = La²+70

Expand

La²+8La+16 = La²+80

8La+16 = 80

8La = 80 - 16

8La = 64

La = 64/8

La = 8cm

Since Lb = La + 4

Lb = 8 + 4

Lb = 12cm

Area of square B = Lb²

Area of square B = 12²

Area of square B = 144cm²

4 0

3 years ago

Read 2 more answers

PLEASE HELP ME FAST!! ( It's about y-intercept)

Alex_Xolod [135]

Answer:

y = 10x + 8

Step-by-step explanation:

1. The slope-intercept form of an equation is y = mx + b, where y = y-coordinate, m = slope, x = x-coordinate, and b = y-intercept.

2. Since we're given the value of slope and y-intercept, we just have to plug in those values to it's corresponding variable.

10 = m, and 8 = b.
y = 10x + 8

6 0

2 years ago

Calculate the area of a circle whose diameter is 156 square

Readme [11.4K]

7.00
Explanation:
Area of a circle
A
=
π
r
2
, where r is the radius.
Given
A
=
154
,
⇒
154
=
π
⋅
r
2

⇒
r
2
=
154
π
=
49.02
⇒
r
=
√
49.02
=
7.00

4 0

3 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

2 years ago

What is 3.14 times 20 time 1/2

ira [324]

Answer:

31.4

Step-by-step explanation:

So a really easy way to solve this is as goes

--> 3.14 x (20 x 1/2)

--> 3.14 x (10)

Now multipling by 10 is just moving the deciaml place one spot to the right

=> thus, we get 31.4

Hope this helps!

5 0

3 years ago