All categories

2 years ago

Am I right please tell me I really NEED IT

Mathematics

1 answer:

AlladinOne [14]2 years ago

5 0

Answer:

it would 13

Step-by-step explanation:

19-6=13 there you go

You might be interested in

If f(x)= |4x - 1|, find the value of x if(x) = 7

vaieri [72.5K]

F(x) = | 4x - 1 | .....when f(x) = 7
7 = | 4x - 1 |

4x - 1 = 7 -(4x - 1) = 7
4x = 7 + 1 -4x + 1 = 7
4x = 8 -4x = 7 - 1
x = 8/4 -4x = 6
x = 2 x = -6/4
x = - 3/2

solutions are x = -3/2, 2

6 0

3 years ago

Read 2 more answers

Simplify.<br><br> 1/4 (1 - 2/3)² + 1/3.<br><br> The answer must be a simplified fraction.

loris [4]

Answer:

$\large\boxed{\dfrac{13}{36}}$

Step-by-step explanation:

Use PEMDAS:

P Parentheses first

E Exponents (ie Powers and Square Roots, etc.)

MD Multiplication and Division (left-to-right)

AS Addition and Subtraction (left-to-right)

---------------------------------------------------------------------------

$\dfrac{1}{4}\left(1-\dfrac{2}{3}\right)^2+\dfrac{1}{3}=\dfrac{1}{4}\left(\dfrac{3}{3}-\dfrac{2}{3}\right)^2+\dfrac{1}{3}=\dfrac{1}{4}\left(\dfrac{3-2}{3}\right)^2+\dfrac{1}{3}\\\\=\dfrac{1}{4}\left(\dfrac{1}{3}\right)^2+\dfrac{1}{3}=\dfrac{1}{4}\cdot\dfrac{1^2}{3^2}+\dfrac{1}{3}=\dfrac{1}{4}\cdot\dfrac{1}{9}+\dfrac{1}{3}=\dfrac{1}{36}+\dfrac{1}{3}\qquad(*)\\\\\text{the common denominator is}\ 36.\\\\36=3\cdot12\to\dfrac{1}{3}=\dfrac{1\cdot12}{3\cdot12}=\dfrac{12}{36}\\\\(*)\qquad=\dfrac{1}{36}+\dfrac{12}{36}=\dfrac{1+12}{36}=\dfrac{13}{36}$

4 0

3 years ago

Which expressions can never result in a negative real number when evaluated for any value of x? Selecta

Y_Kistochka [10]

I think it’s is 5 and take it’s now

8 0

3 years ago

Will mark as BRAINLIEST!!

raketka [301]

Answer:

y - y1 = m (x - x1)

(y - y1) / m = x - x1

x = (y - y1) / m + x1

5 0

3 years ago

the sum of three numbers is 61. the second number is 6 times the first. and the third number is 5 more than the second. find the

mr_godi [17]

$x-the\ first\ number\\6x-the\ second\ number\\6x+5-the\ third\ number\\\\(x)+(6x)+(6x+5)=61\\x+6x+6x+5=61\\13x+5=61\ \ \ \ |subtract\ 5\ from\ both\ sides\\13x=56\ \ \ \ |divide\ both\ sides\ by\ 13\\\boxed{x=\frac{56}{13}\to x=4\frac{4}{13}-the\ first\ number}\\\boxed{6x=6\cdot\frac{56}{13}=\frac{336}{13}=25\frac{11}{13}-the\ second\ number}\\\boxed{6x+5=25\frac{11}{13}+5=30\frac{11}{13}-the\ third\ number}$

6 0

3 years ago