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aleksklad [387]

3 years ago

9

This table shows a proportional relationship between the number of eggs and the number of ounces of tomatoes used for a recipe.

Enter the number of eggs used for 1 ounce of tomato
Eggs Ounces of Tomatoes
4
7
8
14
12
21
Eggs -

1 answer:

laiz [17]3 years ago

6 0

Answer: 12

Step-by-step explanation:

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5(a-2b)-3(a-2b)<br><br>algebra2 .-.

d1i1m1o1n [39]

5(a - 2b) - 3(a - 2b) |use distributive property: a(b - c) = ab - ac

= 5a - 10b - 3a + 6b

= (5a - 3a) + (-10b + 6b)

= 2a - 4b

7 0

3 years ago

Read 2 more answers

Can anybody write an equation for the following word story:

Rudiy27

40=6.95c+5.25
The $40 is what you want it all to equal, so that goes to one side by itself.
The CDs are 6.95 each, so they get a variable to show that.
The 5.95 is added to that, showing that it's only charged once.
Then, all you have to do is solve for c, which is the number of CD's you can buy.

4 0

3 years ago

I need help asap pls and thank you ;)

olga55 [171]

Answer:

$\text{Length of AB is }\frac{ah}{a+h}$

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP (∵ corresponding angles)

∠NDC=∠NPM (∵ corresponding angles)

By AA similarity rule, ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

$\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\$

$KA(\frac{h}{x}-1)=AP$

$ND(\frac{h}{x}-1)=PD$

Adding above two, we get

$(KA+ND)(\frac{h}{x}-1)=(AP+PD)$

⇒ $(KN-AD)=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x^2}{h-x}$

⇒ $x^2=ah-ax-xh+x^2$

⇒ $x(h+a)=ah$

⇒ $x=\frac{ah}{a+h}$

3 0

3 years ago

Please prove it for me

grigory [225]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identity

tanx = $\frac{1}{cotx}$

Consider the left side

$\frac{cotA-1}{cotA+1}$ ← divide terms on numerator/denominator by cotA

= $\frac{\frac{cotA}{cotA}-\frac{1}{cotA} }{\frac{cotA}{cotA}+\frac{1}{cotA} }$

= $\frac{1-tanA}{1+tanA}$

= right side , thus proven

7 0

2 years ago

(12) — (23 + 13i) =<br> Chc

rewona [7]

Answer:

-11 - 13i.

Step-by-step explanation:

(12) — (23 + 13i)

= 12 - 23 - 13i

= -11 - 13i.

4 0

3 years ago