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Alja [10]
3 years ago
15

18. Solve the inequality. Show your work. –6b > 42 or 4b > –4

Mathematics
2 answers:
grigory [225]3 years ago
6 0
-6b > 42 = b<-7
or
4b > -4 = b>-1
Hope this helps.
zhenek [66]3 years ago
4 0

Answer:

Divide each term by -6 and then simplify.    b < -6

divide each term by 2 and then simplify    b > -2

then graph the solution                              b < -6 and b > -2

Step-by-step explanation:


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Ms. Steven creates a drawing in her kitchen. Her drawing is 7cm in length and 4cm in width. The scale on the drawing is 1cm = 3f
marysya [2.9K]

Answer:

7cm=7×3ft=21ft

4cm=4×3ft=12ft

here

length=21ft

breadth=12ft

now

area= l×b

=21ft×12ft

=252ft^2

6 0
3 years ago
Read 2 more answers
HELP<br> Find the area of the polygon with the vertices of A(2,2), B(2,7), C(8,7),and D(4,2)
Illusion [34]

I think it would be 36 it’s what my friend told me I’m not sure though sorry if it’s wrong.

8 0
3 years ago
HELP ME PLZ BRAINLIST 4 U
mylen [45]

Answer:

G

Step-by-step explanation:

I think its G because i counted the numbers...and i pretty sure its not a Negative so yah

8 0
3 years ago
A cash box contains $74 made up of quarters, half-dollars, and one-dollar
Strike441 [17]

Answer:

25 one-dollar coins, 16 half-dollar coins, and 164 quarters

Step-by-step explanation:

First, set up equations based on the information given:

0.25q+0.50h+1.00d=74

\displaystyle{h=\frac{3}{5}d+1}

q=4(d+h)

Then, substitute <em>q</em> in the first equation with the expression from the third equation:

0.25[4(d+h)]+0.50h+1.00d=74\\1d+1h+0.50h+1.00d=74\\2d+1.5h=74

Next, substitute <em>h</em> in that equation with the expression from the second equation:

\displaystyle{2d+1.5(\frac{3}{5}d+1)=74}

2d+0.9d+1.5=74\\2.9d+1.5=74

Solve for <em>d</em>, the number of one-dollar coins:

2.9d+1.5=74\\2.9d=72.5\\d=25

Substitute 25 for <em>d</em> in the second equation to find <em>h</em>, the number of half-dollar coins:

\displaystyle{h=\frac{3}{5}d+1}

\displaystyle{h=\frac{3}{5}(25)+1}

h=15+1\\h=16

Substitute 25 for <em>d</em> and 16 for <em>h</em> in the third equation to find <em>q</em>, the number of quarters:

q=4(d+h)\\q=4(25+16)\\q=4(41)\\q=164

Then, verify that the coins total $74:

0.25(164)+0.50(16)+1.00(25)=74\\41+8+25=74\\74=74\\\text{Check.}

Next, verify that the number of half-dollar coins is one more than three-fifths of the number of one-dollar coins:

\displaystyle{h=\frac{3}{5}d+1}

\displaystyle{16=\frac{3}{5}(25)+1}

16 = 15 + 1\\16 = 16\\\text{Check.}

Finally, verify that the number of quarters is four times the number one-dollar and half-dollar coins together:

q=4(d+h)\\164=4(25+16)\\164=4(41)\\164=164\\\text{Check.}

6 0
3 years ago
Suppose xy=−4 and dy/dt=−3. Find dx/dt when x=−1.
user100 [1]
When x=-1: \quad (-1)y=-4\qquad\to\qquad y=4\)

Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?

x'y+xy'=0

The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.

From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4

and solve for x'.

Hope that helps!
4 0
3 years ago
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