Answer:
<h2>7</h2>
Step-by-step explanation:
![\left[\left(11\:-\:4\right)^3\right]^2\:\div \left(4\:+\:3\right)^5\\\\\frac{\left(\left(11-4\right)^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Subtract\:the\:numbers:}\:11-4=7\\\\=\frac{\left(7^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Add\:the\:numbers:}\:4+3=7\\\\=\frac{\left(7^3\right)^2}{7^5}\\\\\left(7^3\right)^2=7^6\\\\=\frac{7^6}{7^5}\\\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}\\\\\frac{7^6}{7^5}=7^{6-5}\\\\\mathrm{Subtract\:the\:numbers:}\:6-5=1\\\\=7](https://tex.z-dn.net/?f=%5Cleft%5B%5Cleft%2811%5C%3A-%5C%3A4%5Cright%29%5E3%5Cright%5D%5E2%5C%3A%5Cdiv%20%5Cleft%284%5C%3A%2B%5C%3A3%5Cright%29%5E5%5C%5C%5C%5C%5Cfrac%7B%5Cleft%28%5Cleft%2811-4%5Cright%29%5E3%5Cright%29%5E2%7D%7B%5Cleft%284%2B3%5Cright%29%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BSubtract%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A11-4%3D7%5C%5C%5C%5C%3D%5Cfrac%7B%5Cleft%287%5E3%5Cright%29%5E2%7D%7B%5Cleft%284%2B3%5Cright%29%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BAdd%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A4%2B3%3D7%5C%5C%5C%5C%3D%5Cfrac%7B%5Cleft%287%5E3%5Cright%29%5E2%7D%7B7%5E5%7D%5C%5C%5C%5C%5Cleft%287%5E3%5Cright%29%5E2%3D7%5E6%5C%5C%5C%5C%3D%5Cfrac%7B7%5E6%7D%7B7%5E5%7D%5C%5C%5C%5C%5Cmathrm%7BApply%5C%3Aexponent%5C%3Arule%7D%3A%5Cquad%20%5Cfrac%7Bx%5Ea%7D%7Bx%5Eb%7D%3Dx%5E%7Ba-b%7D%5C%5C%5C%5C%5Cfrac%7B7%5E6%7D%7B7%5E5%7D%3D7%5E%7B6-5%7D%5C%5C%5C%5C%5Cmathrm%7BSubtract%5C%3Athe%5C%3Anumbers%3A%7D%5C%3A6-5%3D1%5C%5C%5C%5C%3D7)
Step-by-step explanation:
i don't know but i think this is it
The height is doubled to 8cm to produce a larger box, Box 2 Which phrase accurately describes the volume of Box 2
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°.
Answer:
slope: -11/7
Step-by-step explanation:
Slope is equal to the change in yy over the change in xx, or rise over run.
m=change in ychange in xm=change in ychange in x
The change in xx is equal to the difference in x-coordinates (also called run), and the change in yy is equal to the difference in y-coordinates (also called rise).
m=y2−y1x2−x1m=y2-y1x2-x1
Substitute in the values of xx and yy into the equation to find the slope.
m=19−(8)−9−(−2)m=19-(8)-9-(-2)
Simplify.
m=−11/7
