<em>Given, 9−7x=5−3x</em>
<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.</em>
<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.⇒9−5=7x−3x</em>
<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.⇒9−5=7x−3x⇒4=4x</em>
<em>Given, 9−7x=5−3xPut x terms on one side and constants on another side.⇒9−5=7x−3x⇒4=4x⇒x= </em><em>4</em><em>/</em><em>4</em><em> </em><em> =1</em>
<em> =1Hence, required solution is x=1.</em>
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Answer: x=16
Step-by-step explanation:
For this problem, we can set a proportion to find the side length of the smaller rectangle.
[cross multiply]
[divide both sides by 15]
x=16
1. Start with ΔCIJ.
- ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
- the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
- ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.
2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So
m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.
3. Consider ΔCKL.
- ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
- ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
- the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
- ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.
4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So
m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.
5. ΔABC is isosceles, then angles adjacent to the base are congruent:
m∠KBA=m∠JAB → 222°-8x=205°-7x,
7x-8x=205°-222°,
-x=-17°,
x=17°.
Then m∠CAB=m∠CBA=205°-7x=86°.
Answer: 86°.
1) 10.25 x 10^9 in scientific notation is 1.025 x 10^10
2) Now you have to subtract 10.25x10^9 - 3.0x10^7
To make that you have to put both numbers with the same power of ten.
1025x10^7 - 3.0x10^7 = 1022x10^7 more pennies than half-dollars.
