37-(3×(5+2)-4) please help

Gnom [1K]

9514 1404 393

Answer:

m^2 = 35/9

Step-by-step explanation:

The line will be tangent to the ellipse where there is one point of intersection between the line and the ellipse. We can find that case by looking at the discriminant of the quadratic.

x^2 +9(mx +2)^2 = 1

x^2 +9(m^2x^2 +4mx +4) = 1

(9m^2 +1)x^2+36mx +35 = 0 . . . . subtract 1, collect terms

__

Then the discriminant is ...

(36m)^2 -4(9m^2 +1)(35)

We want it to be zero, so we have ...

1296m^2 -1260m^2 -140 = 0 . . . . eliminate parentheses

36m^2 = 140 . . . . . . . . . . . . . . . add 140

m^2 = 140/36 . . . . . . . . . . . . divide by 36

m^2 = 35/9

3 0

2 years ago

Rewrite the expression with rational exponents as a radical expression. 7 times x to the two thirds power

Daniel [21]

Answer:

$(7x)^{\frac{2}{3}}$ = $\sqrt[3]{49x^{2} }$

Step-by-step explanation:

We have been given the expression;

7 times x to the two thirds power which can be written mathematically as;

$(7x)^{\frac{2}{3}}$

To express the above expression as a radical, we need to recall that;

$a^{\frac{b}{n}}=\sqrt[n]{a^{b}}$

Therefore;

$(7x)^{\frac{2}{3}}=\sqrt[3]{(7x)^{2} }\\\\=\sqrt[3]{49x^{2} }$

6 0

3 years ago

Read 2 more answers

Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit

nignag [31]

Answer:

Step-by-step explanation:

Given that:

$\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\ \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\ where; \\ \\ D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}$

$\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}$

$\text{Now, above equation gives room for double integral in reverse order;}$

$\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy \\ \\= \int^2_0 ( y -0) \ sin (y^2) \ dy \\ \\ = \int^2_0 y \ sin (y^2) \ dy \\ \\ y^2 = U \\ \\ 2y \ dy = du \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du \\ \\ = - \dfrac{1}{2} \Big [cos \ U \Big]^2_o \\ \\ = - \dfrac{1}{2} \Big [cos \ (y^2) \Big]^2_o \\ \\ = - \dfrac{1}{2} cos (4) + \dfrac{1}{2} cos (0) \\ \\$