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3 years ago

What is the degree of the following polynomial? 15x + 9x2 - 6x3 + 1

Mathematics

2 answers:

andrey2020 [161]3 years ago

5 0

Answer:

Step-by-step explanation:

the one have largest power

Crazy boy [7]3 years ago

4 0

253 is the answer I think

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Find each product.

dsp73

Answer:

B) 7v^3-15v^2-17v-3

Step-by-step explanation:

(v-3)(7v^2+6v+1)

7v^3+6v^2+v-21v^2-18v-3

7v^3+6v^2-21v^2+v-18v-3

7v^3-15v^2-17v-3

5 0

3 years ago

Convert to find the equivalent rate. 972 cups minute = quarts minute

bearhunter [10]

Answer:

Convert to find the equivalent rate. 972 cups minute = quarts minute

if there is 4 cups in a quart the answer would be:

972 / 4 = 243

Step-by-step explanation:

4 0

3 years ago

JK⎯⎯⎯⎯⎯ is dilated by a scale factor of n with the origin as the center of dilation, resulting in the image J′K′⎯⎯⎯⎯⎯⎯⎯. The slo

adelina 88 [10]

Its not A but i might be B C D

3 0

3 years ago

Read 2 more answers

What is the value of the expression 10 − ( fraction 1 over 2 )4 ⋅ 48? (1 point)

OlgaM077 [116]

Answer:

Just want to ask if it is 10 - 1/2 × 4 × 48 right? If the expression is this then the answer is -86.

Hope this helps, thank :) !!

6 0

3 years ago

Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1

sleet_krkn [62]

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}$

From Left side:

$\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)$

$\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}$

NOTE: sin²θ + cos²θ = 1

$\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{sin^2\theta-cos^2\theta}$

$\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}$

$\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}$

$\dfrac{2sec^2\theta}{tan^2\theta-1}$

Left side = Right side <em>so proof is complete</em>

8 0

3 years ago

Read 2 more answers