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-Dominant- [34]
3 years ago
14

The area of a rectangular rug is 25.5 square meters. The width of the rug is 3.75 meters. Found the length of the rug by dividin

g its area by its width. What is the length in of the​ rug?
Mathematics
2 answers:
Grace [21]3 years ago
8 0
6.8 because 25.5/3.75 = 6.8
goblinko [34]3 years ago
8 0

Answer: The answer is 6.8 square meters

Step-by-step explanation:

area=25.5 meters

width=3.75 meters

if u want to find the length u should as said divide the area by width\frac{25.5}{3.75}im on school chromebook so i cant see the picture of what i just typed it should say 25.5 over 3.75

one more time:25.5/3.75=6.8

Also i do this in all of my answers now i do something from black panther movie.My favorite line from the movie THIS HOW I BE WHEN I BE ANSWERING QUESTIONS:Okoye:"just dont freeze" Chadwick/Prince(soon to be king;small spoiler) T'challa:What are u talking about..I never freeze. If u watch the movie and say what he said in that accent it will make u think of him and its so fun to do

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Answer: 30y - 25b +5

Step-by-step explanation:

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6 0
3 years ago
Amelia's account balance is $16. If she uses her debit card to buy a $32
zimovet [89]

Answer:

-$16

Step-by-step explanation:

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5 0
3 years ago
Required
svetlana [45]
3.99•10^13 is the answer
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3 years ago
Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure
Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}

Now, we need to express the quadratic polynomials using their roots, as follows:

ay^2+by+c=a(y-y_1)(y-y_2)

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}

Applying the quadratic formula to the second polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the third polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the fourth polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}

Substituting into the rational expression and simplifying:

\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}

8 0
1 year ago
8 × 10 + 6 − 2 7 + 5 × 8 − 4
riadik2000 [5.3K]

Answer:

95

Step-by-step explanation:

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=59+(5)(8)−4

=59+40−4

=99−4

=95

hope this helped :)

6 0
2 years ago
Read 2 more answers
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