All categories

3 years ago

The GCF of 21 and 35 is

Mathematics

1 answer:

finlep [7]3 years ago

4 0

21 = 3 · 7

35 = 5 · 7

You might be interested in

4x+y=10 solve the missing value

nalin [4]

Answer:

Step-by-step explanation:

$4x + y = 10 \\ 4(2) + 2 = 10$

8 0

3 years ago

Estimate the quotient. Round your answer to the nearest whole number.<br><br> 39.8 ÷ 4 ≈ ??

Tresset [83]

Answer:

Step-by-step explanation:

39.8/4=9.95, rounds to 10

5 0

3 years ago

Read 2 more answers

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

3 years ago

Quadrilateral PQRS was translated 5 units to the right and 3 units up to create quadrilateral P′Q′R′S′. Which rule describes thi

Artemon [7]

Answer:

G.(x, y) → (x+5, y+3)

Step-by-step explanation:

The rule would be (x+5, y+3) and the only one with that rule is G.

4 0

3 years ago

Given g(x) = -5x + 5, solve for 2 when g(x) = 0.

yuradex [85]

Answer:

Step-by-step explanation:

g(x) = 0

0=-5x+5

5x=5

x=1

8 0

3 years ago

Read 2 more answers