Ask question

Ask question

All categories

3 years ago

10

Help plzzzzzzzzzzzzzzzz

1 answer:

GenaCL600 [577]3 years ago

6 0

X=6 that is the answer my friend

You might be interested in

Now there is a square city of unknown size with a gate at the center of each side. There is a tree 20 b from the north gate. Tha

Brums [2.3K]

Answer:

The length of each side of the city is 250b

Step-by-step explanation:

Given

$a = 20$ --- tree distance from north gate

$b =14$ --- movement from south gate

$c = 1775$ --- movement in west direction from (b)

See attachment for illustration

Required

Find x

To do this, we have:

$\triangle ADE \sim \triangle ACB$ --- similar triangles

So, we have the following equivalent ratios

$AE:DE = AB:CB$

Where:

$AE = 20\\ DE = x/2 \\ AB = 20 + x + 14 \\ CB = 1775$

Substitute these in the above equation

$20:x/2 = 20 + x + 14: 1775$

$20:x/2 = x + 34: 1775$

Express as fraction

$\frac{20}{x/2} = \frac{x + 34}{1775}$

$\frac{40}{x} = \frac{x + 34}{1775}$

Cross multiply

$x *(x + 34) = 1775 * 40$

Open bracket

$x^2 + 34x = 71000$

Rewrite as:

$x^2 + 34x - 71000 = 0$

Expand

$x^2 + 284x -250x - 71000 = 0$

Factorize

$x(x + 284) -250(x + 284)= 0$

Factor out x + 284

$(x - 250)(x + 284)= 0$

Split

$x - 250 = 0 \ or\ x + 284= 0$

Solve for x

$x = 250 \ or\ x =- 284$

x can't be negative;

So:

$x = 250$

6 0

3 years ago

Passes through (1,4) and is parallel to the line y=2x-3

Novay_Z [31]

Answer:

y = 2x + 2

Step-by-step explanation:

Slope = 2

y-intercept: 4 - (2)(1) = 2

3 0

3 years ago

Factor: 3x^(2)-16x+16

fenix001 [56]

Answer:

$3 {x}^{2} - 16x + 16 \\ 3 {x}^{2} - 12x - 4x + 16 \\ 3x(x - 4) - 4(x - 4) \\ (3x - 4)(x - 4)$

Step-by-step explanation:

I hope it helped U

stay safe stay happy

3 0

3 years ago

Read 2 more answers

Element X decays radioactively with a half-life of eight minutes. If there are 800 g of element X, how long, to the nearest 10th

Harman [31]

Answer: 41.5 min

Step-by-step explanation:

This problem can be solved with the Radioactive Half Life Formula:

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=22 g$ is the final amount of the radioactive element

$A_{o}=800 g$ is the initial amount of the radioactive element

$t$ is the time elapsed

$h=8 min$ is the half life of the radioactive element

So, we need to substitute the given values and find $t$ from (1):

$22 g=(800 g) 2^{\frac{-t}{8 min}}$ (2)

$\frac{22 g}{800 g}=2^{\frac{-t}{8 min}}$ (3)

$\frac{11}{400}=2^{\frac{-t}{8 min}}$ (4)

Applying natural logarithm in both sides:

$ln(\frac{11}{400})=ln(2^{\frac{-t}{8 min}})$ (5)

$-3.593=-\frac{t}{8 min}ln(2)$ (6)

Clearing $t$ :

$t=41.46 min \approx 41.5 min$ This is the time elapsed

7 0

3 years ago

Help please thank you

crimeas [40]

10/30 or you can symlafife

5 0

3 years ago