Ask question

Ask question

All categories

1 year ago

12

Kamal prepares 2 kilograms of dough every hour he works at the bakery. Write an equation that shows the relationship between the

hours worked x and the dough prepared y. *

1 answer:

gogolik [260]1 year ago

4 0

Answer:

<em>2 into x into y</em>

Step-by-step explanation:

You might be interested in

NEED HELP !! There are 16 ounces in 1 pound . How many 12-ounce servings are in 36 pounds of dog food ? PLEASSSEEDE

dedylja [7]

There are (3) 12 ounce servings of dog food

7 0

1 year ago

Read 2 more answers

a paint store has 35 gallons of paint in storage, 2/5 of which are for outdoor use. if each gallon cost $22, what is the total c

Likurg_2 [28]

$462
3/5 of 35 gallons = 21 gallons
21 gallons x $22 = $462

3 0

2 years ago

Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, as shown colon

Nadusha1986 [10]

Answer:

Sadie's error is " she made error in step 2 $=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}$ where $a\geqslant 0$ "

Because she made error in splitting the powers to simplify the square root

<h3>Therefore the correct answer for Sadie's expression is $3ab\sqrt{6ab}$ where $a\geqslant 0$ </h3>

Step-by-step explanation:

Given that " Sadie simplified the expression StartRoot 54 a Superscript 7 b cubed EndRoot, where a greater-than-or-equal-to 0, "

It can be written as $\sqrt{54a^7b^3}$ where $a\geqslant 0$

The given expression is $\sqrt{54a^7b^3}$ where $a\geqslant 0$

To find Sadie's error and explain the correct answer :

Sadie's steps are

$\sqrt{54a^7b^3}$ where $a\geqslant 0$

$=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}$

$=3ab\sqrt{6a^5b}$

<h3> $\sqrt{54a^7b^3}=3ab\sqrt{6a^5b}$ where $a\geqslant 0$ </h3><h3><u>Now corrected steps are</u></h3>

$\sqrt{54a^7b^3}$ where $a≥0$

$=\sqrt{(9\times 6)(a^{6+1})(b^{2+1})$

$=\sqrt{(3^2\times 6)(a^6.a^1)(b^2.b^1)$ (by using the identity $a^{m+n}=a^m.a^n$

$=\sqrt{3^2\times 6\times ((a^3)^2.a)(b^2.b)$ (by using the identity $a^{mn}=(a^m)^n$ )

$=3ab\sqrt{6ab}$

Therefore $\sqrt{54a^7b^3}=3ab\sqrt{6ab}$ where $a\geqslant 0$

<h3>The correct answer is $3ab\sqrt{6ab}$ where $a\geqslant 0$ </h3>

Sadie's error is " she made error in step 2 $=\sqrt{3^2\times 6\times a^2\times a^5\times b^2\times b}$ " where $a\geqslant 0$

Because she made error in splitting the powers to simplify the square root

<h3>Therefore the correct answer for Sadie's expression is $3ab\sqrt{6ab}$ where $a\geqslant 0$ </h3>

5 0

1 year ago

Read 2 more answers

A rectangular pool is 20 feet wide and 50 feet long. A deck used for sunning surrounds the pool. The deck is the same width all

fredd [130]

The answer is B. 3 feet

3 0

1 year ago

part 3. Find the value of the trig function indicated, use Pythagorean theorem to find the third side if you need it.

Nat2105 [25]

Answer: $\bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}$

<u>Step-by-step explanation:</u>

Pythagorean Theorem is: a² + b² = c² , <em>where "c" is the hypotenuse</em>

<em />

$9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}$

Note: 4² + (8√2)² = hypotenuse² → hypotenuse = 12

$10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}$

Note: 12² + opposite² = 20² → opposite = 16

$11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}$

Note: adjacent² + 5² = 6² → adjacent = √11

$12)\ \tan \theta=\dfrac{\text{side opposite of}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{17}{13\sqrt2}\quad =\large\boxed{\dfrac{17\sqrt2}{26}}$

Note: adjacent² + 7² = (13√2)² → adjacent = 17

5 0

1 year ago