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denpristay [2]
3 years ago
10

Solve for x. 14+x>18 Enter your answer, as an inequality, in the box.

Mathematics
1 answer:
Andrej [43]3 years ago
7 0
X is greater than or equal to 5
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Helllllllllllllllllllllllppppppppppp
irinina [24]

Answer:

12.35

Step-by-step explanation:

   95

  *13

____

 285

+950

____

1235

then you need to move the decimal number 2 spaces, because you removed a total of 2 space from the original numbers.

12.35

4 0
3 years ago
Read 2 more answers
X-4=-11 please work problem out it’s homework need help
alisha [4.7K]

X-4=-11

move -4 to the other side

to get X by itself

sign changes from -4 to +4

X-4+4= -11+4

X= -11+4

Answer:

X= -7

5 0
3 years ago
Zair has 48 marbles in his pocket. He has 3 more blues than the
d1i1m1o1n [39]

Answer:

# yellow = 20

# blue = 23

# red = 5

Step-by-step explanation:

let 'r' = # red marbles

let '4r' = # yellow marbles

let '3+4r' = # blue marbles

r + 4r + 3 + 4r = 48

9r + 3 = 48

9r = 45

r = 5

red = 5

yellow = 4(5) = 20

blue = 3+4(5) = 23

6 0
3 years ago
Dr. Newton is saving up to buy a car. She has already saved $3,000. If she wants to spend
Nuetrik [128]

Dr. Newton needs 5000 dollars, and 3000 dollars have already been saved. So, 2000 dollars remain.

If these 2000 dollars must be saved withint 8 months, we can divide the amount needed by the time given, to find out that Dr Newton needs to save

\dfrac{2000}{8}=250

dollars each month.

4 0
3 years ago
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

4 0
2 years ago
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