What is the standard form of y=1x+19

8x+3>_x+10 how do i solve dis<br> HELP

VashaNatasha [74]

Answer:

x > 1

Step-by-step explanation:

Subtract 3 from both sides

8x + 3 - 3 > x + 10 - 3

Simplify

8x > x + 7

Subtract x from both sides

8x - x > x + 7 - x

Simplify

7x - 7

Divide both sides by 7

7x/7 > 7/7

Simplify

x > 1

5 0

3 years ago

kylee bought a pack of 12 cookies. one-third if the cookies are peanut butter. How many of the cookies in the pack are peanut bu

IceJOKER [234]

Answer:

The number of packs are peanuts butter are 4 .

Step-by-step explanation:

As given

Kylee bought a pack of 12 cookies.

one-third if the cookies are peanut butter.

Thus

$Number\ of\ packs\ of\ peanuts\ butter = \frac{1}{3}\times Total\ number\ of\ packs$

$Number\ of\ packs\ of\ peanuts\ butter = \frac{1}{3}\time 12$

$Number\ of\ packs\ of\ peanuts\ butter = \frac{12}{3}$

Number of packs are peanut butter = 4

Therefore the number of packs are peanuts butter are 4 .

4 0

3 years ago

Read 2 more answers

What is the next number is the pattern ?

Lena [83]

18

81 but not sure

25

11

20000

90

6

Am I right?

6 0

3 years ago

A metalworker has a metal alloy that is 15% copper and another alloy that is 75% copper. How many kilograms of each alloy should

attashe74 [19]

X - the mass of the first alloy (15%)
y - the mass of the second alloy (75%)

x kg of the first alloy contains 15%x=0.15x=(15/100)x=(3/20)x kg of copper
y kg of the second alloy contains 75%y=0.75y=(75/100)y=(3/4)y kg of copper
90 kg of a 51% copper alloy contains 51%*90=0.51*90=45.9 kg of copper

$x+y=90 \\
\frac{3}{20}x+\frac{3}{4}y=45.9 \ \ \ |\times (-\frac{4}{3}) \\ \\
x+y=90 \\
\underline{-\frac{1}{5}x-y=-61.2} \\
x-\frac{1}{5}x=90-61.2 \\
\frac{4}{5}x=28.8 \ \ \ |\times \frac{5}{4} \\
x=36 \\ \\
36+y=90 \\
y=90-36 \\
y=54$

The metal worker should combine 36 kg of the 15% copper alloy and 54 kg of the 75% copper alloy to create 90 kilograms of a 51% copper alloy.

7 0

3 years ago

Which equation is the slope-intercept form of the line that passes through (6, –11) and is parallel to the graph of y = –23x + 1

Finger [1]

Answer:

Equation of other line passing through point (6 , - 11) and parallel to given line is y = - 23 x + 127 .

Step-by-step explanation:

Given as :

The equation of one line

y = - 23 x + 12

∵ The equation of line in slope-intercept form is

y = m x + c

where m is the slope of line and c is y-intercept

Now, Compare given line with standard line equation

So, The slope of given line = m = - 23

Now, Again

Other line is passing through point (6 , - 11) and is parallel to given line

so, both the lines are parallel

For parallel line condition , Slope of both lines are equal

Let The slope of other line = M

So, M = m = - 23

Now, Equation of other line passing through point (6 , - 11) and slope - 23 in slope-point form

y - $y_1$ = M × (x - $x_1$ )

i.e y - ( - 11) = (- 23) × (x - 6)

Or, y + 11 = - 23 × x + 138

Or, y = - 23 x + 138- 11

i.e y = - 23 x + 127

So, The equation of other line y = - 23 x + 127

Hence, Equation of other line passing through point (6 , - 11) and parallel to given line is y = - 23 x + 127 . Answer

8 0

3 years ago