All categories

3 years ago

Please help, it’s urgent

Mathematics

1 answer:

WINSTONCH [101]3 years ago

4 0

Answer:

Hey buddy, here is your answer. Hope it helps you.

Step-by-step explanation:

As angle CAB is a linear pair and is 109 degrees. We need to simply subtract it by 180. So 180-109=71. So angle CAE is 71 degrees.

You might be interested in

Convert the following equations from point-slope form to slope-intercept form.

mojhsa [17]

<h2>Answer:</h2>

We need to solve both for y.

#9:

$y - 4 = 3(x - 1)\\\\y - 4 = 3x - 3\\\\y = 3x + 1$

#10:

$y - 1 = -4[x - (-3)]\\\\y - 1 = -4(x + 3)\\\\y - 1 = -4x - 12\\\\y = -4x - 11$

7 0

3 years ago

Becca has a biscuit recipe that uses 2/3 of a tablespoon of salt for every 3/4 of a cup of flour. In order to have enough biscui

MAXImum [283]

2/3   - 3/4
   x    -  3

x= (2/3 * 3) / (3/4) = 2 * (4/3) = 8/3 = 2 and 2/3 of tablespoon of salt

3 0

3 years ago

Read 2 more answers

What is the General form of this equation? (x- -1)^2 +(y-1)^2 =225.0

KATRIN_1 [288]

General Form is the most basic form of an equation and is used as a template to form equations designed to solve a specific problem.

The equation given is "(x- -1)^2 +(y-1)^2 =225.0".

The general form of this Equation would be presented as

·X (+/-) Y = Z
·X^2 (+/-) Y^2 = R1

-I hope this is the answer you are looking for, feel free to post your questions here on brainly in the future.

4 0

3 years ago

30 children are going on a trip it cost £5 including lunch some children take their own packed lunch they pay only £3 the 30 chi

nikitadnepr [17]

No enough info
you can just say that five brought there own lunch while 19 didnt

7 0

4 years ago

Read 2 more answers

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

###

Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

5 0

3 years ago