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3 years ago

5

What are the like terms of 2x^5-3x^3

2 answers:

Kitty [74]3 years ago

7 0

Answer:

There are no like terms

Step-by-step explanation:

The exponents 5 and 3 aren't the same

trasher [3.6K]3 years ago

5 0

Answer:

There are no like terms!

Step-by-step explanation:

Like terms are terms that contain the same variables, and raised to the same power. The numerical coefficients can be different, but the variables AND the exponents must be the same.

In the equation given, 2 and 5 are the coefficients. "x" is the variable, and 5 and 3 are the exponents. Even though the variables are the same, the exponents are not! This means, that each term is separate, and are not like. You cannot combine 2x^5 and 3x^3.

Hope this helps! :)

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Airida [17]

Answer:

20%

Step-by-step explanation:

14/70x100=20%

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3 years ago

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3 years ago

Help! If you know this can you tell me how to do it?

aleksandr82 [10.1K]

Answer:

c

Step-by-step explanation:

Here's how this works:

Get everything together into one fraction by finding the LCD and doing the math. The LCD is sin(x) cos(x). Multiplying that in to each term looks like this:

$[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?$

In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

$\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?$

Put everything over the common denominator now:

$\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?$

Since $sin^2(x)+cos^2(x)=1$ , we will make that substitution:

$\frac{1}{sin(x)cos(x)}$

We could separate that fraction into 2:

$\frac{1}{sin(x)}$ × $\frac{1}{cos(x)}$

$\frac{1}{sin(x)}=csc(x)$ and $\frac{1}{cos(x)}=sec(x)$

Therefore, the simplification is

sec(x)csc(x)

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4 years ago

How is the graph of y=2(3)^x+1 translated from the graph of y = 2(3)^x

umka2103 [35]

y=2(3)x+1

−

y=2(3)x

y-y=2(3)x+1−2(3)x

the answer would be 0=1

but if it is asking u if it is false or true it is

0=1 is false, therefore the system of equations has no solutions

7 0

4 years ago

Find the perimeter of each of the two non congruent triangles where a=15,b=20 and a=29

Alborosie

Answer:

b. about 63.9 units and 41.0 units

Step-by-step explanation:

In question ∠a= 29° and Side of a= 15 and b= 20

Using sine rule of congruence of triangle.

⇒ $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$

⇒ $\frac{15}{Sin 29} = \frac{20}{sin B}$

Using value of sin 29°

⇒ $\frac{15}{0.49} = \frac{20}{sin B}$

Cross multiplying both side.

⇒ Sin B= $\frac{20\times 0.49}{15} = 0.65$

∴ B= 41°

Now, we have the degree for ∠B= 41°.

Next, lets find the ∠C

∵ we know the sum total of angle of triangle is 180°

∴∠A+∠B+∠C= 180°

⇒ $29+41+B= 180$

subtracting both side by 70°

∴∠C= 110°

Now, again using the sine rule to find the side of c.

$\frac{b}{SinB} = \frac{c}{SinC}$

⇒ $\frac{20}{sin41} = \frac{C}{sin110}$

Using the value of sine and cross multiplying both side.

⇒ C= $\frac{20\times 0.94}{0.65} = 28.92$

∴ Side C= 28.92.

Now, finding perimeter of angle of triangle

Perimeter of triangle= a+b+c

Perimeter of triangle= $(15+20+28.92)= 63.9$

∴ Perimeter of triangle= 63.9 units

7 0

3 years ago