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mafiozo [28]
3 years ago
7

What is the value of x?Enter your answer in the box.x=​

Mathematics
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

x = 8

Step-by-step explanation:

This is an equilateral triangle, so we know that the angles are all equal too.

That means each angle is 60 degrees (b/c 60 * 3 = 180)

7x + 4 = 60

7x = 56

x = 8

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A man gathers nuts in a forest. Each time he finds a nut, there is a 20% chance it is a pecan and a 35% chance it is a walnut.
gladu [14]
Either a pecan or walnut means you add the probabilities together

20% plus 35% equals 55%
7 0
3 years ago
Toni and Marcy partnered together in running laps around a track to raise money for the Children's Hospital's playroom. Toni rai
lutik1710 [3]
X represent the number of laps Marcy swam, 114x represent the number of times Toni swam.

20 + 0.80(114x) + 15 + 0.85x = 257
20 + 91.2x + 15 + 0.85x = 257
92.05x + 35 = 257

Required equation is 92.05x + 35 = 257
3 0
2 years ago
I hurry i need help asap
anzhelika [568]
It is the first chart
6 0
2 years ago
A detective finds a victim of murder at 9 am. At that time, the temperature of the body is 90.3 degrees F. One hour later, the b
LekaFEV [45]

Answer:

Step-by-step explanation:

Given

at 9 am temperature of the body is 90.3^{\circ}F

1 hr later temperature 89^{\circ}F

Temperature of room T=68^{\circ}F

According to newtons law

\frac{\mathrm{d} T}{\mathrm{d} t}=-k\left ( T-T_{room}\right )

\frac{\mathrm{d} T}{\mathrm{d} t}=-k\left ( T-68\right )

\frac{dT}{T-68}=-kdt

Integrating

=\int \frac{dT}{T-68}=-kdt

\ln |T-68|=-kt+c

T-68=Ae^{-kt}

Now at 9 am i.e. t=0,

90.3=A+68

A=22.3

At 10 am i.e. t=1 hr T=89

89-68=22.3e^{-k}

e^{-k}=21

k=0.06006

Temperature of Normal human body is T=98.6 ^{\circ}F

98.6-68=e^{-0.06006t}

30.6=e^{-0.06006t}

Taking log both sides

t=-5.27

i.e. 5 hr and  16.2 min

i.e. victim murdered at 3 am and 43.8 min

7 0
3 years ago
X=square root of 880-8x
Sunny_sXe [5.5K]
x=\sqrt{880-8x}\\
D:880-8x\geq0 \wedge x\geq0\\
D:8x\leq880 \wedge x\geq0\\
D:x\leq 110 \wedge x\geq0\\
D:x\in\langle0,110\rangle\\
x^2=880-8x\\
x^2+8x-880=0\\
x^2+8x+16-896=0\\
(x+4)^2=896\\
x+4=8\sqrt{14} \vee x+4=-8\sqrt{14}\\
x=-4+8\sqrt{14} \vee x=-4-8\sqrt{14}\\
\\ x=-4-8\sqrt{14}\not \in D \Rightarrow \boxed{x=-4+8\sqrt{14}}
5 0
3 years ago
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