Answer:
5.95
Step-by-step explanation:
Five and ninety-five hundredths as a decimal number is 5.95.
Hoped this helped.
The answer is 8! Hope that helps! Please give brainliest hehe
Hi! Your answer is q = -9
Please see an explanation for a better and clear understanding to your problem.
Any questions about my answer and explanation can be asked through comments! :)
Step-by-step explanation:
Since we want to solve for q-term. That means we are going to isolate q-term.
![\huge{3q+4+9=-14}](https://tex.z-dn.net/?f=%5Chuge%7B3q%2B4%2B9%3D-14%7D)
We can add 4 and 9 together.
![\huge{3q+13=-14}](https://tex.z-dn.net/?f=%5Chuge%7B3q%2B13%3D-14%7D)
Because we want to know the value of q. That means we have to isolate q-term by subtracting both sides by 13.
![\huge{3q+13-13=-14-13}\\\huge{3q=-27}](https://tex.z-dn.net/?f=%5Chuge%7B3q%2B13-13%3D-14-13%7D%5C%5C%5Chuge%7B3q%3D-27%7D)
We are reaching to the final step where we divide the whole equation by 3.
![\huge{\frac{3q}{3}=-\frac{27}{3}}\\\huge{q=-9}](https://tex.z-dn.net/?f=%5Chuge%7B%5Cfrac%7B3q%7D%7B3%7D%3D-%5Cfrac%7B27%7D%7B3%7D%7D%5C%5C%5Chuge%7Bq%3D-9%7D)
Finally, the solution for this equation is q = -9. But what if you are not certain or sure about the answer? Let's check it out!
To check the answer, simply substitute q = -9 in the equation.
![\huge{3q+4+9=-14}\\\huge{3(-9)+13=-14}\\\huge{-27+13=-14}\\\huge{-14=-14}](https://tex.z-dn.net/?f=%5Chuge%7B3q%2B4%2B9%3D-14%7D%5C%5C%5Chuge%7B3%28-9%29%2B13%3D-14%7D%5C%5C%5Chuge%7B-27%2B13%3D-14%7D%5C%5C%5Chuge%7B-14%3D-14%7D)
Notice that the equation is true for q = -9. Hence, we can conclude that the solution for this equation is q = -9.
Hope this helps!
Answer:
mÐ4 = 30°
Step-by-step explanation:
Vertical angles are angles that are opposite to each other, thus are said to be equal.
Since,
i. Ð1 and Ð3 are vertical angles
ii. Ð2 and Ð4 are vertical angles
This implies that the measure of Ð1 and Ð3 are congruent. And also the measure of Ð2 and Ð4 are congruent.
So that, if mÐ2 = 30°, it would be expected that mÐ4 has the same measure. This implies that mÐ4 = 30°.
Ooh, fun
geometric sequences can be represented as
![a_n=a(r)^{n-1}](https://tex.z-dn.net/?f=a_n%3Da%28r%29%5E%7Bn-1%7D)
so the first 3 terms are
![a_1=a](https://tex.z-dn.net/?f=a_1%3Da)
![a_2=a(r)](https://tex.z-dn.net/?f=a_2%3Da%28r%29)
![a_2=a(r)^2](https://tex.z-dn.net/?f=a_2%3Da%28r%29%5E2)
the sum is -7/10
![\frac{-7}{10}=a+ar+ar^2](https://tex.z-dn.net/?f=%5Cfrac%7B-7%7D%7B10%7D%3Da%2Bar%2Bar%5E2)
and their product is -1/125
![\frac{-1}{125}=(a)(ar)(ar^2)=a^3r^3=(ar)^3](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B125%7D%3D%28a%29%28ar%29%28ar%5E2%29%3Da%5E3r%5E3%3D%28ar%29%5E3)
from the 2nd equation we can take the cube root of both sides to get
![\frac{-1}{5}=ar](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B5%7D%3Dar)
note that a=ar/r and ar²=(ar)r
so now rewrite 1st equation as
![\frac{-7}{10}=\frac{ar}{r}+ar+(ar)r](https://tex.z-dn.net/?f=%5Cfrac%7B-7%7D%7B10%7D%3D%5Cfrac%7Bar%7D%7Br%7D%2Bar%2B%28ar%29r)
subsituting -1/5 for ar
![\frac{-7}{10}=\frac{\frac{-1}{5}}{r}+\frac{-1}{5}+(\frac{-1}{5})r](https://tex.z-dn.net/?f=%5Cfrac%7B-7%7D%7B10%7D%3D%5Cfrac%7B%5Cfrac%7B-1%7D%7B5%7D%7D%7Br%7D%2B%5Cfrac%7B-1%7D%7B5%7D%2B%28%5Cfrac%7B-1%7D%7B5%7D%29r)
which simplifies to
![\frac{-7}{10}=\frac{-1}{5r}+\frac{-1}{5}+\frac{-r}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B-7%7D%7B10%7D%3D%5Cfrac%7B-1%7D%7B5r%7D%2B%5Cfrac%7B-1%7D%7B5%7D%2B%5Cfrac%7B-r%7D%7B5%7D)
multiply both sides by 10r
-7r=-2-2r-2r²
add (2r²+2r+2) to both sides
2r²-5r+2=0
solve using quadratic formula
for
![ax^2+bx+c=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%2Bc%3D0)
![x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
so
for 2r²-5r+2=0
a=2
b=-5
c=2
![r=\frac{-(-5) \pm \sqrt{(-5)^2-4(2)(2)}}{2(2)}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B-%28-5%29%20%5Cpm%20%5Csqrt%7B%28-5%29%5E2-4%282%29%282%29%7D%7D%7B2%282%29%7D)
![r=\frac{5 \pm \sqrt{25-16}}{4}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5%20%5Cpm%20%5Csqrt%7B25-16%7D%7D%7B4%7D)
![r=\frac{5 \pm \sqrt{9}}{4}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5%20%5Cpm%20%5Csqrt%7B9%7D%7D%7B4%7D)
![r=\frac{5 \pm 3}{4}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5%20%5Cpm%203%7D%7B4%7D)
so
![r=\frac{5+3}{4}=\frac{8}{4}=2](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5%2B3%7D%7B4%7D%3D%5Cfrac%7B8%7D%7B4%7D%3D2)
or
![r=\frac{5-3}{4}=\frac{2}{4}=\frac{1}{2}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B5-3%7D%7B4%7D%3D%5Cfrac%7B2%7D%7B4%7D%3D%5Cfrac%7B1%7D%7B2%7D)
use them to solve for the value of a
![\frac{-1}{5}=ar](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B5%7D%3Dar)
![\frac{-1}{5r}=a](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B5r%7D%3Da)
try for r=2 and 1/2
![a=\frac{-1}{10}](https://tex.z-dn.net/?f=%20a%3D%5Cfrac%7B-1%7D%7B10%7D)
or
![a=\frac{-2}{5}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B-2%7D%7B5%7D)
test each
for a=-1/10 and r=2
a+ar+ar²=
![\frac{-1}{10}+\frac{-2}{10}+\frac{-4}{10}=\frac{-7}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B10%7D%2B%5Cfrac%7B-2%7D%7B10%7D%2B%5Cfrac%7B-4%7D%7B10%7D%3D%5Cfrac%7B-7%7D%7B10%7D)
it works
for a=-2/5 and r=1/2
a+ar+ar²=
![\frac{-2}{5}+\frac{-1}{5}+\frac{-1}{10}=\frac{-7}{10}](https://tex.z-dn.net/?f=%5Cfrac%7B-2%7D%7B5%7D%2B%5Cfrac%7B-1%7D%7B5%7D%2B%5Cfrac%7B-1%7D%7B10%7D%3D%5Cfrac%7B-7%7D%7B10%7D)
it works
both have the same terms but one is simplified
the 3 numbers are
![\frac{-2}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B-2%7D%7B5%7D)
,
![\frac{-1}{5}](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B5%7D)
, and