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solniwko [45]
2 years ago
11

V=ZQT Solve for z in this equation

Mathematics
1 answer:
Ganezh [65]2 years ago
4 0

Answer:

Z= V/QT

Step-by-step explanation:

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Help with question 10.
kogti [31]
C.add10 is the correct answer
6 0
2 years ago
Show work and explain with formulas:
Marina86 [1]

17 Answer:  205

<u>Step-by-step explanation:</u>

\{1\dfrac{2}{3}+1\dfrac{5}{6}+2+...+8\dfrac{1}{3}\}\implies a_1=1\dfrac{2}{3},\ d=\dfrac{1}{6}\\\\\\a_n=a_1+d(n-1)\qquad solve\ for\ n\\\\8\dfrac{1}{3}=1\dfrac{2}{3}+\dfrac{1}{6}(n-1)\\\\\\\dfrac{25}{3}=\dfrac{5}{3}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{50}{6}=\dfrac{10}{6}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{41}{6}=\dfrac{1}{6}n\\\\\\\dfrac{41}{6}\cdot 6=n\\\\41=n

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{41}=\dfrac{1\dfrac{2}{3}+8\dfrac{1}{3}}{2}\cdot 41\\\\\\.\quad =\dfrac{10}{2}\cdot 41\\\\\\.\quad =5\cdot 41\\\\.\quad =\large\boxed{205}

18 Answer:  1968

<u>Step-by-step explanation:</u>

a_1=-6,\ d=2,\ n=48, \quad \text{solve for }a_{48}\\\\a_{n}=a_1+d(n-1)\\\\a_{48}=-6+2(48-1)\\\\.\quad =-6+2(47)\\\\.\quad =-6+94\\\\.\quad =88

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{48}=\dfrac{-6+88}{2}\cdot 48\\\\\\.\quad =\dfrac{82}{2}\cdot 48\\\\\\.\quad =41\cdot 48\\\\.\quad =\large\boxed{1968}

19 Answer:  -116

<u>Step-by-step explanation:</u>

\{3, -2, -7, ...\}\\a_1=3,\ d=-5,\ n=8, \quad \text{solve for }a_{8}\\\\a_{n}=a_1+d(n-1)\\\\a_{8}=3-5(8-1)\\\\.\quad =3-5(7)\\\\.\quad =3-35\\\\.\quad =-32\\\\\text{Now use the sum formula:}\\\\S_8=\dfrac{a_1+a_8}{2}\cdot 8\\\\\\S_{8}=\dfrac{3-32}{2}\cdot 8\\\\\\.\quad =\dfrac{-29}{2}\cdot 8\\\\\\.\quad =-29\cdot 4\\\\.\quad =\large\boxed{-116}

6 0
2 years ago
PLEASE HELP ME EASYYYYYY MATH QUESTION BRAINLIEST
Burka [1]

Answer:

8% is equivalent to the fraction 8/100

Step-by-step explanation:

7 0
2 years ago
Write a quadratic equation given the roots -1/3 and 5, show your work
Travka [436]

\boxed{(x - a)(x - b) = 0}

The equation above is the intercept form. Both a-term and b-term are the roots of equation.

x =  -  \frac{1}{3}  \\ x = 5

These are the roots of equation. Therefore we substitute a = - 1/3 and b = 5 in the equation.

(x +  \frac{1}{3} )(x -  5) = 0

Here we can convert the expression x+1/3 to this.

x +  \frac{1}{3}  = 0 \\  3x + 1 = 0

Rewrite the equation.

(3x + 1)(x - 5) = 0

Simplify by multiplying both expressions.

3 {x}^{2}  - 15x + x - 5 = 0 \\ 3 {x}^{2}  - 14x  - 5 = 0

<u>Answer</u><u> </u><u>Check</u>

Substitute the given roots in the equation.

3 {(5)}^{2}  - 14(5)  - 5 = 0 \\ 75 - 70 - 5 = 0 \\ 75 - 75 = 0 \\ 0 = 0

3( -  \frac{1}{3} )^{2}  - 14( -  \frac{1}{3}) - 5 = 0 \\ 3( \frac{1}{9} ) +  \frac{14}{3}  - 5 = 0 \\  \frac{1}{3}  +  \frac{14}{3}  -  \frac{15}{3}  = 0 \\  \frac{15}{3}  -  \frac{15}{3}  = 0 \\ 0 = 0

The equation is true for both roots.

<u>Answer</u>

\large \boxed {3 {x}^{2}  - 14x - 5 = 0}

8 0
2 years ago
Please help with this question !!
mario62 [17]

Answer:

ha

Step-by-step explanation:

ha

3 0
2 years ago
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