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3 years ago

Drag each expression to show whether the product is less than 1

Mathematics

1 answer:

barxatty [35]3 years ago

3 0

Equal to 1:

$3*\frac{1}{3}\ \ , \ \ 5*\frac{1}{5}\ \ , \ \6*\frac{1}{6}$

Less than 1:

$1*\frac{5}{6}$

Greater than 1:

$2 * \frac{3}{4} \ \ , \ \ \ 3 * \frac{2}{5}$

Step-by-step explanation:

We will solve each product one by one.

So,

$3*\frac{1}{3} = 1\\\\6*\frac{1}{6} = 1\\\\2 * \frac{3}{4} = \frac{3}{2}\\\\1*\frac{5}{6} = \frac{5}{6}\\\\3*\frac{2}{5} = \frac{6}{5}\\\\5*\frac{1}{5} = \frac{1}{5}$

As we have all the answers now.

If the denominator of a fraction is greater than ts numerator then the result is less than 1
If the denominator of a fraction is less than its numerator then the result is greater than 1.

So,

Equal to 1:

$3*\frac{1}{3}\ \ , \ \ 5*\frac{1}{5}\ \ , \ \6*\frac{1}{6}$

Less than 1:

$1*\frac{5}{6}$

Greater than 1:

$2 * \frac{3}{4} \ \ , \ \ \ 3 * \frac{2}{5}$

Keywords: Fractions, decimals

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tensa zangetsu [6.8K]

Answer:

(6t) + (3t) + (4t)

Simplified it would be 13t

3 0

3 years ago

Distance between the points

Harman [31]

To understand the distance formula, you first need to understand the Pythagorean Theorem. For a refresher, the theorem states that the square of the legs of a right triangle is equal to the the square of its hypotenuse (the side opposite the right angle), or in symbols:

$a^2+b^2=c^2$ , where a and b are the lengths of the legs, and c is the length of the hypotenuse. In the context of the x-y plane, the legs of the triangle correspond to separate x and y values on the plane, and the hypotenuse corresponds to a straight line between two points on that plane.

To find the distance between the points you've listed, (2√5,4) and (1,2√3), we'll first need to find the "legs" of the triangle. To find the length of the x leg, we'll just need the distance between the x values of the points, which we find to be 2√5-1. We do the same for the y component, which ends up being 4-2√3. Now that we have our legs, we're ready to find the hypotenuse - or the distance.

Going back to Pythagorus's equation, we have:

$(2 \sqrt{5}-1)^{2}+(4-2 \sqrt{3})^{2}=d^2$

where d, the hypotenuse of the triangle, means "distance."

To solve for d, we take the square root of both sides:

$d= \sqrt{(2 \sqrt{5}-1)^2+(4-2 \sqrt{3} )^2}$

And from there, all that's left to do is solve the right side of the equation, which just ends up being rote calculation.

Edit: I'll go through the steps of that calculation here. We'll start by expanding each of the squared terms inside the radical:

$(2 \sqrt{5}-1)^2=(2 \sqrt{5}-1)(2 \sqrt{5}-1)=(2 \sqrt{5}-1)2 \sqrt{5}-(2 \sqrt{5}-1)$
$=(2 \sqrt{5})^2-2 \sqrt{5}-2 \sqrt{5}+1=20-4\sqrt{5}+1$

$(4-2\sqrt{3})^2=(4-2\sqrt{3})(4-2\sqrt{3})=(4-2\sqrt{3})4-(4-2\sqrt{3})2\sqrt{3}$
$=16-8\sqrt{3}-8\sqrt{3}+(2\sqrt{3})^2=16-16\sqrt{3}+12$

Putting those values back under the radical:

$\sqrt{20-4\sqrt{5}+1+16-16\sqrt{3}+12}$

Collecting constants:

$\sqrt{49-4\sqrt{5}-16\sqrt{3}}$

If you wanted an exact answer, this messy-looking thing would be it, and you can verify those results on WolframAlpha if you'd like. If you want an approximation, just enter that expression in to the online calculator of your choice, and it should give out the value of approx. <span>3.51325.</span>

In general, if you want to solve for the distance between two points $(y_{1},x_{1})$ and $(y_{2},x_{2})$ , the formula is:

$d= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

4 0

3 years ago

Need help ASAP I tried to do the problem but I don't know how

inysia [295]

Answer:

see below

Step-by-step explanation:

(a) the graph is symmetrical about the horizontal axis. It has a maximum value of r = 3 at θ = π.

(b) The graph is symmetrical about both the horizontal and the vertical axis. It has a maximum value of r = 1 at θ = 0 and θ = π. (Note this curve has subtle differences from the inner loop of the above curve.)

_____

<em>About how</em>

As with any graphing problem, when doing it by hand, one chooses enough points to give the general shape of the curve. In some cases, quite a few points may be required. It is often helpful to use a spreadsheet for calculating the point values. Here, we've graphed the equations using "technology"--a graphing calculator.