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4 years ago

Solve plz and zoom in if can't see

Mathematics

1 answer:

DiKsa [7]4 years ago

8 0

6(3x+8)+32+12x
6•3x+6•8+32+12x
18x+48+32+12x
30x+48+32
30x/80
0.375

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FREEEEEEEEE POINTSSSSSSSSS

noname [10]

Answer:

sick ty

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In parallelogram mnpq, side mn is seven inches longer than twice the length of side np if the perimeter of mnpq is 68 inches the

frozen [14]

Perimeter of a parallelogram = 2(a+b)
Perimeter of parallelogram mnpq = 68 inches
<span>mn is seven inches longer than twice the length of side np
let the length of np = x
length of mn= 2x + 7
Now,
p = 2(2x+7+x)
68 = 4x + 14 + 2x
68 = 6x + 14
6x = 68 - 14
6x = 54
x = 9
Thus, the length of np is 9 inches.

</span>

8 0

4 years ago

Read 2 more answers

Q1 : Gina writes the letters of the word PROBABILITY on separate cards. She picks one card at random. What is the probability th

disa [49]

Answer:

2/11

4/11

6/11

Step-by-step explanation:

probability the letter B = number of times b appears in the word / total number of alphabets

total number of alphabets = 11

number of times b appears in the word = 2

2/11

a vowel = total number of vowels in the word / total number of alphabets

vowels include = a,e,i,o,u

vowels in the word = a, o, i

total number of vowels in the word = 4

4/11

A letter before M in the alphabet = total number of alphabets before m in the word / total number of alphabets

letters before m = A, B, C, D, E, F, G, H, I, J, K, L

letters before m in the word = b,a, i , l

number of words before m = 6/11

4 0

3 years ago

In the diagram, AB is a tangent to the circle, centre O. D is the mid-point of the chord BC. Given that BAC = x, find COD in the

Westkost [7]

Answer:

$\angle COD =\frac{90+x}{2}$ in term of x

Step-by-step explanation:

Given that AB is tangent to a circle with center O and radius of OC=OB

D is the mid-point of the chord BC and D is 90

Here, Angle BAC = x.

From figure,

AC is a straight line.

we can write as

$\angle AOB+\angle BOD +\angle COD =180$

Since D is the mid-point of the chord BC

We know,

Angle opposite to sides are congruent

$\angle BOD=\angle COD$

So,

$\angle AOB+\angle BOD +\angle COD =180$

$\angle AOB +2\angle COD =180$

Now, In triangle AOB

$\angle ABO = 90$

Therefore,

$\angle AOB+\angle ABO+\angle BAO=180$

$\angle AOB=90-x$

Therefore.

$\angle AOB +2\angle COD =180$

$(90-x) +2\angle COD =180$

$-x+2\angle COD =90$

$2\angle COD =90+x$

$\angle COD =\frac{90+x}{2}$

Thus, $\angle COD =\frac{90+x}{2}$

6 0

4 years ago

Find the solutions of <img src="https://tex.z-dn.net/?f=y%3Dx%5E%7B2%7D%2B2" id="TexFormula1" title="y=x^{2}+2" alt="y=x^{2}+2"

Eva8 [605]

Answer:

x is $\frac{1\ + \sqrt{3} i}{2}$ AND

$\frac{1\ - \sqrt{3} i}{2}$

y = $\frac{3\ + \sqrt{3} i}{2}$ And

$\frac{3\ - \sqrt{3} i}{2}$

Step-by-step explanation:

Given two equations are :

y = x² + 2 And

y = x + 1

So, the equation can be written as

x² + 2 = x + 1

Or, x² - x + ( 2 - 1) = 0

Or, x² - x + 1 = 0

This is in the form of quadratic equation

So, Roots of equation x be :

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

Or, x = $\frac{1\pm \sqrt{-1^{2}-4\times 1\times 1}}{2\times 1}$

Or , x = $\frac{1\pm \sqrt{-3}}{2}$

Hence the two value of x is $\frac{1\ + \sqrt{3} i}{2}$ AND

$\frac{1\ - \sqrt{3} i}{2}$

So , y = $\frac{3\ + \sqrt{3} i}{2}$ And

$\frac{3\ - \sqrt{3} i}{2}$

5 0

3 years ago