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3 years ago

I don't understand this problem can someone please help me

Mathematics

1 answer:

Stella [2.4K]3 years ago

5 0

ok kid you need to study instead of getting people to give you the answer

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Difference of two squares Calculate this 81-(3x-2)^2 ^2= to the power of 2

Marat540 [252]

81-(3x-2)^2
= 81-(-6)^2
= 81 - 36
= 45

hope it helps

5 0

3 years ago

A baker is making croissants. He has 18 pounds of dough. Each croissant is made from 1/6 pounds of dough. How many croissants ca

Black_prince [1.1K]

Answer:The baker can make 144 croissants

Step-by-step explanation:

Lets solve your question

Divide the total amount of lbs of dough (18 lbs) by the total amount of dough for each croissant (1/8 lbs)

First divide 18(lbs) by 1/8 (lbs)

18/1/8= 18*8=144 croissants

7 0

2 years ago

Angle BAC measures 56°.

wel

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Do you know how to solve this (2x + 4)(x + 3)

nasty-shy [4]

Answer:

18x

Step-by-step explanation:

get the x by itself then it is 2+4=6 so that is 6x and then do 3+x=3x and 6x×3x= 18x

7 0

3 years ago

A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra

Harman [31]

Answer:

A.

The student made an error in step 3 because a is positive in Quadrant IV; therefore,

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

Step-by-step explanation:

Given

$P\ (a,b)$

$r = \± \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}$

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

$r = \sqrt{(a)^2 + (b)^2}$

Since a belongs to the x axis and b belongs to the y axis;

$cos\theta$ is calculated as thus

$cos\theta = \frac{a}{r}$

Substitute $r = \sqrt{(a)^2 + (b)^2}$

$cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}$

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}}$

Rationalize the denominator

$cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}$

$cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}$

So, from the list of given options;

The student's mistake is that a is positive in quadrant iv and his error is in step 3

3 0

3 years ago