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2 years ago

Solve 12.7 ≥ s − 5.3 Graph the solution.

Mathematics

Guest

2 years ago

i had the egsat same question

1 answer:

Julli [10]2 years ago

8 0

Hello!

I dont a graph to show you but I will solve the answer.

12.7 + 5.3 ≥ s

18 ≥ s

s ≤ 18
That is your answer.

Enjoy.
~Isabella

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From 2000-2003, students were tested by the state in four major subject areas of math, science, English and social studies. The

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2 years ago

What is 10 times 7 thousands in unit form

Irina18 [472]

10 times 7 is 70.
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2 years ago

Read 2 more answers

Given that $A = (\sqrt{2008}+\sqrt{2009}),$ $B = (-\sqrt{2008}-\sqrt{2009}),$ $C = (\sqrt{2008}-\sqrt{2009}),$ and $D = (\sqrt{2

kipiarov [429]

Answer:

ABCD = 0.803

Step-by-step explanation:

$A = \sqrt{2008}+\sqrt{2009} \\A = 44.81 + 44.82\\A = 89.63$

$B = -\sqrt{2008}-\sqrt{2009}\\B = -44.81 - 44.82\\B = -89.63$

$C = \sqrt{2008}-\sqrt{2009}\\C = 44.81 - 44.82\\C = - 0.01$

$D = \sqrt{2009}-\sqrt{2008}\\D = 44.82 - 44.81\\D = 0.01$

ABCD = 89.63 * (-89.63) * (-0.01) * (0.01)

ABCD = 0.803

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2 years ago

Jordan solved the equation −7x + 25 = 48; his work is shown below. Identify the error and where it was made.

Mashcka [7]

The answer is C, Step 3: He should have divided both sides by -7.

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2 years ago

Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an

Charra [1.4K]

1. Consider square ABCD. You know that

$A_{ABCD}=AD^2=200,$

then

$AB=BC=CD=AD=10\sqrt{2}.$

2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that

$FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.$

3. Consider trapezoid BFGC. Its area is

$A_{BFGC}=\dfrac{FG+BC}{2}\cdot h,$ where h is the height of trapezoid and is equal to half of AB. Thus,

$A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.$

$A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.$

5. Note that angles EGC and CGD are supplementary and

$\sin \angle CGD=\sin \angle EGC.$

Then

$A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.$

Answer: $A_{CGD}=41.$

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2 years ago