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blondinia [14]
3 years ago
9

Which number line shows the solution to zrc - 5 > 3?

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
7 0

Answer:

4th option

Step-by-step explanation:

⅔x - 5 > 3

⅔x > 8

x > 8 × 3/2

x > 12

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HELP! PLZ I NEED THIS ASAP
harina [27]

Answer:

2x^{4}+x^{3}-x+2

Step-by-step explanation:

To find which one is prime, let's try to factor them all. We can use the factoring by grouping method.

x^{3}+3x^{2}-2x-6

x^{3}+3x^{2}and-2x-6

x^{2}(x+3)and-2(x+3)

So this one is not prime, since you can still factor it.

x^{3}+2x^{2}-3x-6

x^{3}+2x^{2}and-3x-6

x^{2}(x-2)and3(x-2)

So this one is not prime, since you can still factor it.

4x^{4}+4x^{2}-2x-2

4x^{4}+4x^{2}and-2x-2

4x^{3}(x+1)and-2(x+1)

So this one is not prime, since you can still factor it.

2x^{4}+x^{3}-x+2

2x^{4}+x^{3}and-x+2

x^{3}(2x+1)and -x+2 cannot be further factored.

Therefore, 2x^{4}+x^{3}-x+2 is a prime.

6 0
4 years ago
Can someone please do number 30.
Mamont248 [21]
To\ find\ b\ you\ have\ to\ put\ (2,6)\ into\ second\ eqution:\\\\y=2.5x+b\\\\6=2.5\cdot2+b\\6=5+b\\6-5=b\\b=1\\\\b\ have\ to\ be\ equal\ 1.
6 0
4 years ago
If x is an acute angle and tan x = 5,
lyudmila [28]
We have that
tan x=5

we know that
tan x =sin x/ cos x
sin x/ cos x=5------> squaring-----> sin² x/cos² x=25
remember that
sin² x+cos² x=1-----> cos² x=1-sin² x
substitute
sin² x/[1-sin² x]=25
sin² x=25-25*sin² x
26*sin² x=25
sin² x=25/26
sin x=5/√26

cos² x=1-(5/√26)²----> 1-(25/26)----> 1/26
cos x=1/√26

the answers are
sin x=5/√26
cos x=1/√26

4 0
3 years ago
A circle has a diameter of 2.5 ft.
alukav5142 [94]
A = pi * r^2
D = 2.5 so r = 1.25
A = 3.14 * 1.25^2
A = 4.9 ft^2
8 0
3 years ago
Whose good at Geomtry, specifically trigonometry?!!?
ddd [48]

Answer:

C. 2.8 miles per minute

Step-by-step explanation:

The mnemonic SOH CAH TOA reminds you that ...

... Tan = Opposite/Adjacent

In the relevant triangle, the side opposite the angle at the observer is the altitude of the airplane, 2 miles. The side adjacent is the horizontal distance to the airplane. At the first observation, that distance (d1) is ...

... tan(40°) = (2 mi)/d1

At the second observation, the horizontal distance to the airplane (d2) is ...

... tan(50°) = (2 mi)/d2

Solving for d1 and d2 and finding the difference (∆d), we have ...

... d1 = (2 mi)/tan(40°)

... d2 = (2 mi)/tan(50°)

... ∆d = d1 -d2 = (2 mi)(1/tan(40°) -1/tan(50°) ≈ 2·(1.1918 -0.8391) mi

... ∆d ≈ 2°0.3526 mi ≈ 0.7053 mi

This distance was flown by the plane in 15 seconds, so it will travel 4 times this distance in 60 seconds (1 minute).

... ∆d/∆t = (0.7053 mi)/(1/4 min) = 4·0.7053 mi/min ≈ 2.8 mi/min

7 0
4 years ago
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