Ask question

Ask question

All categories

Yakvenalex [24]

3 years ago

11

Solve for b. -1lb + 7 = 40

1 answer:

MariettaO [177]3 years ago

7 0

Answer:

b = -33/l

Step-by-step explanation:

-1lb + 7 = 40

-1lb = 33 (subtraction property of equality)

-1lb/l = 33/l (division property of equality)

-1b/-1 = -33/l (division property of equality)

b = -33/l (solved)

You might be interested in

SOMEONE PLEASE ANSWER ASAP FOR BRAINLIEST!!!!!

kaheart [24]

Answer:

D b = -2E +212

Step-by-step explanation:

The slope is given by -2 for every 1000 ft

The initial value is 212 degrees

We can use

y = mx+b where m is the slope and b is the initial value

y = -2x +212

We are at an elevation of E thousand ft

Replace x with E

y = -2 *E +212

b is the boiling point at this elevation

We replace y with b

b = -2E +212

7 0

3 years ago

While driving past stores, Tatum counted the number of cars in the parking lots.

Vlad [161]

Answer:

9

Step-by-step explanation:

1, 3, 5, 9, 15, 19, 19

7 0

3 years ago

Read 2 more answers

What is (x+2)(x+5) in polynomial standard form

Harlamova29_29 [7]

Answer:

=x^2+7x+10

Step-by-step explanation:

Simplify: xx+5x+2x+2.5: x+7x+10

Add similar elements: 5x+2x= 7x

=xx+7x+2.5

xx=x2

2.5=10

=x+7x+10

Answer: =x^2+7x+10

<em><u>Hope this helps.</u></em>

3 0

3 years ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

3 0

3 years ago

PLS HELP!!! due to inflation nominal prices are usually there were to demand increases. After the second increase the price for

SashulF [63]

Answer:

The first increase was of 60%.

Step-by-step explanation:

The initial value of the product is x.

The first increase was of y.

The second increase is of 25%, that is, 1.25.

The final price was double the original, so 2x.

This situation can be modeled by the following equation:

$xy(1.25) = 2x$

We want to find y.

Simplifying by x

$1.25y = 2$

$y = \frac{2}{1.25}$

$y = 1.6$

After the first increase, the value was 1.6 of the original value, that is a increase as a percent of (1.6 - 1)*100 = 60%.

8 0

3 years ago