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malfutka [58]
3 years ago
7

Anyone know what 13+(x+8) is

Mathematics
1 answer:
babymother [125]3 years ago
6 0
Answer:
13 + x + 8
= x + 21
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Match the following items by evaluating the expression for
stich3 [128]

Answer:

x^-2 = 1/4

x^-1 = -1/2

x^0  = 1

x^1 = -2

x^2 = 4

Step-by-step explanation:

got it right

5 0
3 years ago
What is the least common denominator of1/3 and 1/5
ella [17]
The LCD is 1/15 and 3/15.
8 0
3 years ago
Read 2 more answers
Dose anyone know how to solve this?
sesenic [268]

Answer:

AC is double the length of AB

so AC = 12

Step-by-step explanation:

mark me as brainliest ❤️

8 0
3 years ago
Given: ∆PQR, m∠R = 90° m∠PQR = 75° M ∈ PR , MP = 18 m∠MQR = 60° Find: RQ
tensa zangetsu [6.8K]

Answer:    RQ= 8.99 ( approx)

Step-by-step explanation:

Let MR= x

Since, In triangle, PRQ, tan 75°= \frac{18+x}{RQ}

⇒ RQ=  \frac{18+x}{tan 75^{\circ}}

Now, In triangle MRQ,

tan 60°= \frac{18+x}{RQ}

⇒ RQ=  \frac{x}{tan 60^{\circ}}

On equating both values of RQ,

\frac{18+x}{tan 75^{\circ}}=\frac{x}{tan 60^{\circ}}

⇒\frac{18+x}{x}=\frac{tan 75^{\circ}}{tan 60^{\circ}}

⇒\frac{18+x}{x}=\frac{tan 75^{\circ}}{tan 60^{\circ}}

⇒\frac{18+x}{x}=2.15470053838

⇒18=2.15470053838x-x

⇒x=15.5884572681≈15.60

Thus RQ=8.99999999999≈8.99


6 0
3 years ago
Suppose that the position of one particle at time is given by x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π and the position of a second p
Mashcka [7]

Answer:

there is no collision between the particles

Step-by-step explanation:

for the first particle

x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π

for the second particle

x2 = -3 + cos t, y2 = 1 + sin t, 0 ≤ t ≤ 2π

then for the collision

x1=x2 → 3*sin t = -3 + cos t → sin t= -1 + (cos t)/3→ 1+ sin t = (1/3)cos t  

y1=y2 → 1 + sin t = 2 cos t → (1/3)cos t  = 2 cos t →(1/3) = 2

since 1/3 ≠ 2 there is no collision between the particles

6 0
3 years ago
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