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Pachacha [2.7K]
2 years ago
7

yall help meee with this work

Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx" id="TexFormula1" title=" x^{2} \geq \lim_{n \to \infty} {ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \\ \int\limits^a_b {x} \, dx" alt=" x^{2} \geq \lim_{n \to \infty} {ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \\ \int\limits^a_b {x} \, dx" align="absmiddle" class="latex-formula">

Mathematics
1 answer:
Andreyy892 years ago
8 0

Answer:

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Step-by-step explanation:

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HELP MEEE???? L'LL GIVE U A BRAINLIST???
Katarina [22]
Okay, don’t freak out. It’s more simple than it seems.
To find the area of this rectangle you use length x length
Let’s look for the length first
Use the bottom left and right points
(-2.25,-1) and (2.25,-1)
The length is a horizontal line so we use the x-coordinates. The distance from -2.25 to 2.25 is 4.5
Now let’s find the height. Let’s use the top left and bottom left coordinates.
(-2.25,4) and (-2.25,-1)
The height is a vertical line. The y-coordinates. Their difference is 5.
So 4.5 x 5 = 22.5
3 0
2 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
Through: (3, 4), slope = -2​
sineoko [7]

Answer:

y = -2x - 2

Step-by-step explanation:

Slope-Intercept Form: y = mx + b

Points: (3, 4)

Slope: -2

y = -2x + b

4 = -2 ( 3 ) + b

4 = -6 + b

b = -2

y = -2x - 2

8 0
2 years ago
Read 2 more answers
Select the two values of x that are roots of this equation.
alina1380 [7]
C
;$3&;!/‘rnjakfhenfbwmdbx
4 0
3 years ago
A 30 gram sample of a substance that's used to sterilize surgical instruments has a k-value of 0.1235. Find the substances half
jok3333 [9.3K]

Answer : The substances half life in days is, 5.6 days.

Step-by-step explanation :

Half life : It is defined as the amount of time taken by a radioactive material to decay to half of its original value.

All radioactive decays follow first order kinetics.

The relation between the half-life and rate constant is:

k=\frac{0.693}{t_{1/2}}

where,

k = rate constant = 0.1235 per days

t_{1/2} = half-life

Now put all the given values in the above formula, we get:

0.1235\text {days}^{-1}=\frac{0.693}{t_{1/2}}

t_{1/2}=5.6\text{ days}

Thus, the substances half life in days is, 5.6 days.

5 0
3 years ago
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