Ask question

Ask question

All categories

3 years ago

13

The smallest of three consecutive integers is added to four times the largest integer; the result so obtained is 15 less than th

ree times the middle integer. Find the integers.

1 answer:

Anton [14]3 years ago

3 0

Answer:

Let the three consecutive integers be x,x+1,x+2

ATQ

x+2(x+2)+15=4(x+1)

=3x+19=4x+4

= x=15

..... Hence. The nos. Are 15,15+1=16 and 15+2=17

Step-by-step explanation:

You might be interested in

There are twelve people on the jury there were 4more man than women how many man were <br>there/

irina1246 [14]

$x-the \ number \ of \ women \\ x+4 -the \ number \ of \ men \\\\ x+x+4=12 \\\\ 2x+4=12 \\\\ 2x=12-4 \\\\ 2x=8 \\\\ \boxed{x=\frac{8}{2}=4 \ women} \\\\ x+4= \\\\ = 4+4\to\boxed{8 \ men}$

4 0

3 years ago

Read 2 more answers

what is the expression, equation, or inequality using x as your variable: The product of a number and its opposite, increased by

olchik [2.2K]

$x\cdot (-x)+7$

7 0

3 years ago

Help on functions please.

dedylja [7]

A) Sub g(x) and h(x) into the equation:
x^2 + 3x - 40 - (-x -3)
= x^2 + 3x - 40 +x + 3
= x^2 + 4x - 37
Find x:
x= 4.40 and x= -8.40
b) Sub f(x) and g(x) into the equation:
(x^2 - 64) / (x^2 + 3x -40)
= ((x+8) x (x-8)) / ((x+8) x (x-5))
= (x - 8) / (x-5)
Find x:
x= 8 and x = 5
How to get those number:
(x^2 - 64) = x^2 - 8^2 = (x-8) x (x+8)
( x^2 + 3x - 40) = (x^2 + +8x - 5x - 40) = x( x + 8) - 5( x + 8) = (x-5) x (x+8)

Try to do c and d :)

3 0

4 years ago

Unit 8: Right Triangles & Trigonometry<br> Homework 2: Special Right Triangles

Vsevolod [243]

Answer:

Part 1) $x=14\sqrt{2}\ units$

Part 2) $y=14\ units$

Part 3) $z=14\sqrt{3}\ units$

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ABC

we know that

The triangle ABC is a $45^o-90^o-45^o$

so

Is an isosceles right triangle

The legs are equal

therefore

AC=AB

$x=14\sqrt{2}\ units$

step 2

Find the length side BC

Applying the Pythagorean Theorem]

$BC^2=AB^2+AC^2$

$BC^2=(14\sqrt{2})^2+(14\sqrt{2})^2$

$BC^2=784\\BC=28\ units$

step 3

Find the value of y

In the right triangle BCD

$sin(30^o)=\frac{BD}{BC}$ ----> by SOH (opposite side divided by the hypotenuse)

$sin(30^o)=\frac{y}{28}$

Remember that

$sin(30^o)=\frac{1}{2}$

so

$\frac{y}{28}=\frac{1}{2}\\\\y=14\ units$

step 4

Find the value of z

In the right triangle BCD

$cos(30^o)=\frac{DC}{BC}$ ----> by CAH (adjacent side divided by the hypotenuse)

$cos(30^o)=\frac{z}{28}$

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

so

$\frac{z}{28}=\frac{\sqrt{3}}{2}\\z=14\sqrt{3}\ units$

3 0

3 years ago

Read 2 more answers

I need help answering these 1. And 2. <br> 10 points ⭐️<br> If helped

ratelena [41]

Well they are not in length

4 0

4 years ago

Read 2 more answers