All categories

3 years ago

Test test tes t testtest

Mathematics

2 answers:

algol [13]3 years ago

6 0

What what what what what what what

yuradex [85]3 years ago

4 0

..............................................

You might be interested in

How many digits will be to the right of the decimal point in the product? 1.456 × 84

Monica [59]

Decimal pt after 3 digits in first number so 3 digits in product.

6 0

3 years ago

Read 2 more answers

In a quiz, positive marks are given for correct answers and negative marks are given for incorrect answers. If Muskan's scores i

Elanso [62]

Answer:

Step-by-step explanation:

To find his total score in the quiz, we need to add all the scores he had in successive rounds.

This gives a value of; 25 + (-5) + (-10) + 15 + 10 = 25-5-10+15+10 = 25 + 10 = 35

Thus his total score at the end is 35

3 0

3 years ago

Suppose y varies directly with x, and y = 12 when x = 2. What direct variation

neonofarm [45]

Answer:

$y=24$

Step-by-step explanation:

Given:

$y=kx$

Substitute:

$\left \{ {{y=12} \atop {x=2}} \right.$ into $y=kx$

$12=k\times2$

Swap the sides:

$k\times2=12$

Divide both sides of the equation by the coefficient of variable:

$k=\frac{12}{2}$

Cross out the common factor:

$k=6$

Find the equation of the function:

$y=6x$

Substitute:

$y=6\times4$

Calculate the product or quotient:

$y=24$

4 0

2 years ago

Find r(t) if r'(t) = 5t4i + 6t5j + t k and r(1) = i + j.

Vladimir [108]

$\mathbf r(t)=\displaystyle\int(5t^4\,\mathbf i+6t^5\,\mathbf j+t\,\mathbf k)\,\mathrm dt$
$\mathbf r(t)=(t^5+C_1)\,\mathbf i+(t^6+C_2)\,\mathbf j+\left(\dfrac12t^2+C_3\right)\,\mathbf k$

Given $\mathbf r(1)=\mathbf i+\mathbf j$ , we have

$\begin{cases}1^5+C_1=1\\\\1^6+C_2=1\\\\\dfrac121^2+C_3=0\end{cases}\implies C_1=0,C_2=0,C_3=-\dfrac12$

so that the particular solution is

$\mathbf r(t)=t^5\,\mathbf i+t^6\,\mathbf j+\dfrac12(t^2-1)\,\mathbf k$

8 0

3 years ago

Show work and explain with formulas:

Marina86 [1]

17 Answer: 205

Step-by-step explanation:

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

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

18 Answer: 1968

Step-by-step explanation:

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

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

19 Answer: -116

Step-by-step explanation:

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

6 0

3 years ago