All categories

3 years ago

Find the value of x. 6 14 3x-12 A o 21 x = [?] Enter

Mathematics

2 answers:

liq [111]3 years ago

8 0

Answer:

Bueno esto es lo que yo calculo

3×‐12

3(×‐4)

3 14x12

A calcule

A 12×3 =14

es lo que me calcula

Anton [14]3 years ago

3 0

Answer:

Step-by-step explanation:

14/6 = 21 / 3x - 12

first, cross multiply

14 (3x - 12) = 6*21

Remove the brackets

42x - 168 = 126

Add 168 to both sides

42x = 168 + 126

Combine

42x = 294

Divide by 42

x = 294/42

You might be interested in

Can someone help me with problem 6 and 7? Please explain I am confused

scZoUnD [109]

This is interesting. You cannot turn 1/4 or 2/4 or 3/4 into a mixed number because they are proper fractions. Perhaps ask your teacher about it? Or just rewrite the fractions as is and state they are in proper form.

Same with 7. They are each simplified already. I would suggest asking your teacher.

7 0

3 years ago

One number is seven less than four times the other number. The sum of the two numbers is 93. Find the numbers.

ivann1987 [24]

You can write the two equations x = 4y - 7 and x + y = 93; substitute the value of x into the other equation

(4y - 7) + y = 93; combine like terms

5y - 7 = 93; add 7 to both sides

5y = 100; divide both sides by 5

y = 20; substitute this value into the other equation

x + 20 = 93

x = 73

So the numbers are 20 and 73

6 0

3 years ago

Round 141.999 to the nearest tenth hundredth ten and hundred

astraxan [27]

141.999 rounded to the nearest
tenth = 142
hundreth =142
ten = 140
hundred = 100

4 0

3 years ago

I need help with this,I will mark braily to the right answer!

pentagon [3]

Answer:

where the qutsion

Step-by-step explanation:

3 0

3 years ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

Read 2 more answers