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3 years ago

Which is greater? 4 tenths and 2 hundredths or 2 tenths and 4 hundredths

Mathematics

1 answer:

Orlov [11]3 years ago

8 0

4 tenths and 2 hundredths is written .42

2 tenths and 4 hundreds is written .24

.42 is greater than .24

.42 > .24

Hope this helps.

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A parallelogram can have one pair or two pairs of parallel sides

Rudiy27

Answer:

A parallelogram can have 2 pairs of parallel sides.

Step-by-step explanation:

Think of a rectangle for a parallelogram. It has 2 pairs of parallel sides which makes it a parallelogram. The same for a square as well as it has 2 parallel pairs. Just remember that a parallelogram always has 2 pairs of parallel sides.

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3 years ago

CAN SOMEONE GOOD AT MATH HELP PLEEESE!!

Rufina [12.5K]

Answer: 1 .Thus for a graph to have an Euler circuit, all vertices must have even degree. The converse is also true: if all the vertices of a graph have even degree, then the graph has an Euler circuit, and if there are exactly two vertices with odd degree, the graph has an Euler path.

2. A graph has an Euler circuit if and only if the degree of every vertex is even. A graph has an Euler path if and only if there are at most two vertices with odd degree.

Step-by-step explanation:

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2 years ago

What is the answer to. 4x + 7<8

sukhopar [10]

First, subtract 7 on both sides. Now, you have 4x<1. Divide by 4 on both sides. Finally, you get x<1/4 (a quarter or said as one-fourth). So, x<1/4.

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2 years ago

Match the identities to their values taking these conditions into consideration sinx=sqrt2 /2 cosy=-1/2 angle x is in the first

BaLLatris [955]

Answer:

$\cos(x+y)$ goes with $-\frac{\sqrt{6}+\sqrt{2}}{4}$

$\sin(x+y)$ goes with $\frac{\sqrt{6}-\sqrt{2}}{4}$

$\tan(x+y)$ goes with $\sqrt{3}-2$

Step-by-step explanation:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

We are given:

$\sin(x)=\frac{\sqrt{2}}{2}$ which if we look at the unit circle we should see

$\cos(x)=\frac{\sqrt{2}}{2}$ .

We are also given:

$\cos(y)=\frac{-1}{2}$ which if we look the unit circle we should see

$\sin(y)=\frac{\sqrt{3}}{2}$ .

Apply both of these given to:

$\cos(x+y)$

$\cos(x)\cos(y)-\sin(x)\sin(y)$ by the addition identity for cosine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}-\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2}$

$\frac{-\sqrt{2}}{4}-\frac{\sqrt{6}}{4}$

$\frac{-\sqrt{2}-\sqrt{6}}{4}$

$-\frac{\sqrt{6}+\sqrt{2}}{4}$

Apply both of the givens to:

$\sin(x+y)$

$\sin(x)\cos(y)+\sin(y)\cos(x)$ by addition identity for sine.

$\frac{\sqrt{2}}{2}\frac{-1}{2}+\frac{\sqrt{3}}{2}\frac{\sqrt{2}}{2}$

$\frac{-\sqrt{2}+\sqrt{6}}{4}$

$\frac{\sqrt{6}-\sqrt{2}}{4}$

Now I'm going to apply what 2 things we got previously to:

$\tan(x+y)$

$\frac{\sin(x+y)}{\cos(x+y)}$ by quotient identity for tangent

$\frac{\sqrt{6}-\sqrt{2}}{-(\sqrt{6}+\sqrt{2})}$

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}$

Multiply top and bottom by bottom's conjugate.

When you multiply conjugates you just have to multiply first and last.

That is if you have something like (a-b)(a+b) then this is equal to a^2-b^2.

$-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}-\sqrt{2}}$

$-\frac{6-\sqrt{2}\sqrt{6}-\sqrt{2}\sqrt{6}+2}{6-2}$

$-\frac{8-2\sqrt{12}}{4}$

There is a perfect square in 12, 4.

$-\frac{8-2\sqrt{4}\sqrt{3}}{4}$

$-\frac{8-2(2)\sqrt{3}}{4}$

$-\frac{8-4\sqrt{3}}{4}$

Divide top and bottom by 4 to reduce fraction:

$-\frac{2-\sqrt{3}}{1}$

$-(2-\sqrt{3})$

Distribute:

$\sqrt{3}-2$

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2 years ago

Y=2x. 1x=-y+6<br> Solve this system of equations

Allushta [10]

Answer:

y=2

x=1

Step-by-step explanation:

check the image attached

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2 years ago