christine baked 55 with 11 scoups of flour, how many scoups of flour does christine need in order to bake 70 cookies

Please help me its a math problem thank you i give brainlist

inna [77]

Answer:

1.875

Step-by-step explanation:

the average rate of change of g(x) in the closed interval [ a, b ] is

$\frac{g(b)-g(a)}{b-a}$

here [a, b ] = [- 1, 3 ] , then

g(b) = g(3) = 2³ + 1 = 8 + 1 = 9

g(a) = g(- 1) = $2^{-1}$ + 1 = $\frac{1}{2}$ + 1 = 1.5

then

average rate of change = $\frac{9-1.5}{3-(-1)}$ = $\frac{7.5}{4}$ = 1.875

8 0

3 years ago

Read 2 more answers

Since YouTube first became available to the public in mid-2005, the rate at which video has been uploaded to the site can be app

Sergeeva-Olga [200]

$\int\limits^7_1 {525,600(0.42t^2+2.7t-1)} \, dx \\=525,600 \left \{\frac{0.42t^3}{3}+\frac{2.7t^2}{2}-t \right. \left \} {{7} \atop {1}} \right. \\=525,600\{[0.14(343)+1.35(49)-7]-[0.14+1.35-1]\}\\=525,600(107.17-0.49)\\=525,600(106.68)\\=56,071,008$
Total number of video uploaded is 56 million

7 0

3 years ago

A Fruit Stand sells 6 oranges for $3 and 3 grapefruits for $2.40 Lionel buys 10 oranges and 11 grapefruit how much Lionel spend

lina2011 [118]

$3/6 = .50 per orange
$2.40/3 = .80 per grapefruit

10(.50) + 11(.80) =
$5.00 + $8.80 = $13.80

8 0

4 years ago

Read 2 more answers

Hayley is considering two gym membership plans. The plans are detailed here.

Novosadov [1.4K]

Option 2: $y = 3x+50$ is the right answer

Step-by-step explanation:

We have to convert the details given in Plan one to an equation

Let x be the number of visits to the gym

and y be the total cost for plan one

So,

As each visit costs $3, x number of visits will cost 3x

One time fee is $50

So the equation can be written as:

$y = 3x+50$

Hence

Option 2: $y = 3x+50$ is the right answer

Keywords: Slope equations of lines

Learn more about slope at:

#LearnwithBrainly

5 0

3 years ago

Prove: cot(x) sec^4(x) = cot(x) +2tan(x) + tan^3(x)

masha68 [24]

$\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\qquad\qquad 
sec(\theta)=\cfrac{1}{cos(\theta)}\quad\qquad 1+tan^2(\theta)=sec^2(\theta)\\\\\\
tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}
\qquad \qquad cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\\\\
-------------------------------\\\\
cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)\\\\
-------------------------------\\\\$

$\bf \textit{so, let's do the left-hand-side}\\\\
cot(x)sec^2(x)sec^2(x)\implies cot(x)[1+tan^2(x)][1+tan^2(x)]
\\\\\\
cot(x)[1^2+2tan^2(x)+tan^4(x)]
\\\\\\
cot(x)+2tan^2(x)cot(x)+tan^4(x)cot(x)
\\\\\\
cot(x)+2\cdot \cfrac{sin^2(x)}{cos^2(x)}\cdot \cfrac{cos(x)}{sin(x)}+\cfrac{sin^4(x)}{cos^4(x)}\cdot \cfrac{cos(x)}{sin(x)}
\\\\\\
cot(x)+2\cdot \cfrac{sin(x)}{cos(x)}+\cfrac{sin^3(x)}{cos^3(x)}\implies \boxed{cot(x)+2tan(x)+tan^3(x)}$

8 0

3 years ago