Which equation is equivalent to the equation 6x+9=12?

ABCD is a rectangle. Its diagonals meet at O. Find the value of x if BO = 3x + 4 andOC = x + 8 .Also find the length of its diag

scoundrel [369]

Step-by-step explanation:

the lengths OB, OC & BC together forms a isosceles triangle.

therefore,

OB=OC

3x+4=x+8

2x=4

x=2.

OB = 3x+4= 3.2+4=10cm.

OC= x+8= 2+8=10cm.

therefore, the diagonals are : 20cm each.

hope this helps you.

3 0

2 years ago

Now find the components nx and ny of n⃗ in the tilted coordinate system of part b.

Norma-Jean [14]

solution:

vector N points to the left and up so the components Nx and Ny are negative and positive respectively

then ,

Nx = - N sinθ and Ny = N cosθ

3 0

3 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).

Nezavi [6.7K]

Answer:

The ratio of cows to pigs can be expressed as 3:5

The ratio of horses to chickens can be expressed as 3:4

The ratio of sheep to the total number of animals can be expressed as 7:5

Step-by-step explanation:

I might be wrong

8 0

3 years ago

In ΔDEF, the measure of ∠F=90°, the measure of ∠D=19°, and DE = 40 feet. Find the length of EF to the nearest tenth of a foot.

mars1129 [50]

Answer:

13.0feet

Step-by-step explanation:

Given ΔDEF with ∠F=90°, the measure of ∠D=19°, and DE = 40 feet, since one of the angles of the triangle is 90°, the triangle is a right angled triangle as shown in the attachment.

Using SOH, CAH, TOA to get the unknown EF.

Since <D is opposite to side EF, EF will be the opposite while side DE will be the hypotenuse

Based on SOH:

Sin<D = Opposite/Hypotenuse

Sin<D = EF/DE

Sin19° = EF/40

EF = 40sin19°

EF = 13.0 feet to the nearest tenth of a foot

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=4x2%20%2B%20x%20-%203" id="TexFormula1" title="4x2 + x - 3" alt="4x2 + x - 3" align="absmiddle

Mnenie [13.5K]

Answer:

$\boxed{\textsf{ The roots of the quadratic equation is \textbf{ (-4) and 3 }.}}$

Step-by-step explanation:

A quadratic equation is given to us . Here we aren't given what to do with it. So , lets find out the roots of the equation . The given quadratic equation is :-

$\sf\implies 4x^2+x-3 =0$

Now with respect to Standard form ax²+bx + c . The roots of the equation can be find out by the quadratic formula also known as " Shreedharyacharya's Formula " .

Quadratic Formula :-

$\boxed{\boxed{\sf x =\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}}$

Now with respect to Standard Form we have ,

• a = 4 •b = 1 •c = (-3)

Put on the respective values :-

$\sf\implies x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a} \\\\\sf\implies x =\dfrac{-1\pm\sqrt{1^2-4(1)(-3)}}{2(1)}\\\\\sf\implies x =\dfrac{-1\pm \sqrt{1+48}}{2}\\\\\sf\implies x =\dfrac{-1\pm \sqrt{49}}{2}\\\\\sf\implies x = \dfrac{-1\pm 7 }{2}\\\\\sf\implies x =\dfrac{-1-7}{2},\dfrac{-1+7}{2}\\\\\sf\implies x =\dfrac{-8}{2},\dfrac{6}{2}\\\\\sf\implies \boxed{\pink{\sf x = (-4) , 3 }}$

7 0

3 years ago