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3 years ago

14

Pls help mehhhhh!!!!!!!!!

1 answer:

OlgaM077 [116]3 years ago

6 0

Answer:

Domain = all real numbers

Range = {-1 [greater than or equal to] y [greater than] Infinity}

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Solve the equation using the zero-product property. (2x − 8)(7x + 5) = 0 x = –2 or x = 7 x = 4 or x = x = 4 or x = x = –4 or x =

aliya0001 [1]

$(2x-8)(7x+5)=0\iff2x-8=0\ \vee\ 7x+5=0\\\\2x-8=0\qquad\text{add 8 to both sides}\\2x=8\qquad\text{divide both sides by 2}\\\boxed{x=4}\\\\7x+5=0\qquad\text{subtract 5 from both sides}\\7x=-5\qquad\text{divide both sides by 7}\\\boxed{x=-\dfrac{5}{7}}$

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3 years ago

Helen rented a kayak at the lake. There is a charge of $6 as an initial fee and then $3 an hour for every hour that she uses the

Helga [31]

Helen kayaked for 5 hours.
Hope this helps!!

6 0

3 years ago

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The answer is p = -2/3 and q= -2

igomit [66]

$y=\dfrac{p}{q+x}\\\\\dfrac{1}{y}=\dfrac{1}{\frac{p}{q+x}}=\dfrac{q+x}{p}$

We have x and y intercept:

(2, 0) and (0, 3).

Substitute the coordinates of points to the equation:

$(2,\ 0)\\\\0=\dfrac{q+2}{p}\to q+2=0\to q=-2\\\\(0,\ 3)\\3=\dfrac{q+0}{p}\to3p=q\to3p=-2\to p=-\dfrac{2}{3}$

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3 years ago

Evaluate the line integral, where C is the given curve. (x + 6y) dx + x2 dy, C C consists of line segments from (0, 0) to (6, 1)

Dima020 [189]

Split $C$ into two component segments, $C_1$ and $C_2$ , parameterized by

$\mathbf r_1(t)=(1-t)(0,0)+t(6,1)=(6t,t)$

$\mathbf r_2(t)=(1-t)(6,1)+t(7,0)=(6+t,1-t)$

respectively, with $0\le t\le1$ , where $\mathbf r_i(t)=(x(t),y(t))$ .

We have

$\mathrm d\mathbf r_1=(6,1)\,\mathrm dt$

$\mathrm d\mathbf r_2=(1,-1)\,\mathrm dt$

where $\mathrm d\mathbf r_i=\left(\dfrac{\mathrm dx}{\mathrm dt},\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt$

so the line integral becomes

$\displaystyle\int_C(x+6y)\,\mathrm dx+x^2\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}(x+6y,x^2)\cdot(\mathrm dx,\mathrm dy)$

$=\displaystyle\int_0^1(6t+6t,(6t)^2)\cdot(6,1)\,\mathrm dt+\int_0^1((6+t)+6(1-t),(6+t)^2)\cdot(1,-1)\,\mathrm dt$

$=\displaystyle\int_0^1(35t^2+55t-24)\,\mathrm dt=\frac{91}6$

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3 years ago

MATH HELP !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

sveta [45]

So the first one, 1.2 >11/5. and the second one 1/3> 1/2

6 0

3 years ago

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