All categories

4 years ago

Is -1/3 - 4/5 positive, negative, or zero?

Mathematics

1 answer:

nata0808 [166]4 years ago

6 0

It is negative,

Hope this helped.

You might be interested in

Find y' by implicit differentiation √x +√y =1

AleksandrR [38]

Differentiate both sides with respect to x

$\frac{d}{dx}(x^{\frac{1}{2}} + y^{\frac{1}{2}}) = \frac{d}{dx}(1)\\\\ \frac{1}{2x^{\frac{1}{2}}} + \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0\\\\\frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}\\\\\frac{1}{2\sqrt{y}}\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \\\\\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \times 2\sqrt{y}\\\\\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \\\\\boxed{\bf{y'= -\frac{\sqrt{y}}{\sqrt{x}}}}$

5 0

3 years ago

To solve a problem, we often _________ the given information into algebraic expressions and equations.

vovangra [49]

We often substitute the given info

4 0

3 years ago

Which expression is equivalent to 2⁵

Misha Larkins [42]

Answer:

2*2*2*2*2

Step-by-step explanation:

You just multiply 2 five times.

7 0

3 years ago

A football team lost 9 yards on its first play. It then gained 9 yards on its second play.

MrMuchimi

Answer:

Your Answer Should Be C It’s the hade that most sense and if you read carefully you can the right to the question! ✍

C. Suppose that, instead, the football team had gained 9 yards on its first play and lost 9 yards on its second play. Would the team’s position on the field have been any different than it actually was after the first two plays? Why or why not?

5 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

VladimirAG [237]

Answer:

a = 5.

Step-by-step explanation:

The graph here is a probability density graph. What will represent $P(X\le a)$ on the graph?

$P(X\le a)$ is the area

between the graph and the x-axis,
to the left of $a$ .

The area under the graph between 0 and 9 is a trapezoid. Consider the trapezoid in three slices from left to right:

A right triangle of area $\displaystyle \frac{1}{2}\times 0.2\times 4 = 0.4$ ,
A rectangle of area $0.2 \times (5 - 4) = 0.2$ , and
Another right triangle of area $\displaystyle \frac{1}{2} \times 0.2 \times (9 - 5) = 0.4$ .

The area of the leftmost triangle plus that of the rectangle is exactly 0.6. In other words, the area to the left of $a = 5$ between the graph and the x-axis is $0.6$ . $P(X \le 5) = 0.6$ . $a = 5$ .

As a side note $a = 5$ shall be the only answer to this question since the area under the graph to the left of $a$ can only increase or stay constant but not decrease as the value of $a$ increases.

3 0

3 years ago