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prisoha [69]
3 years ago
14

18 tens 20 ones = ______hundreds

Mathematics
1 answer:
Elis [28]3 years ago
3 0

Answer:

200

Step-by-step explanation:

18 × 10 = 180

20 × 1 = 20

180 + 20 = 200

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Which statement is true regarding the functions on the
nordsb [41]

Answer: C

Step-by-step explanation:

intersection of f(x) and g(x) is x=-3

f(-3)=g(-3)

Answer C

8 0
2 years ago
Neptune’s average distance from the sun is 4.503 x 10e9 km. Mercury’s average distance from the sun is 5.791 x10e9 km.about how
Igoryamba

Answer:

77.76 times

Step-by-step explanation:

The average distance of Neptune  from the sun

= 4.503  ×  10 ⁹  k m .

and Mercury  =  5.791  ×  10 ⁷ k m .

Hence neptune is (  4.503  ×  10 ⁹) ÷ (5.791 × 10 ⁷  ) times farther from the sun than mercury

i.e.(  \frac{4.503}{5.791} )  × 10⁹⁻⁷ times

=   0.7776  ×  10 ² times

=   77.76  times.

4 0
3 years ago
In ΔRST, t = 4.1 inches, r = 7.1 inches and ∠S=45°. Find the length of s, to the nearest 10th of an inch.
dem82 [27]

Answer:

The length of s is 5.1 inches to the nearest tenth of an inch

Step-by-step explanation:

In Δ RST

∵ t is the opposite side to ∠T

∵ r is the opposite side to ∠R

∵ s is the opposite side to ∠S

→ To find s let us use the cosine rule

∴ s² = t² + r² - 2 × t × r × cos∠S

∵ t = 4.1 inches, r = 7.1 inches, and m∠S = 45°

→ Substitute them in the rule above

∴ s² = (4.1)² + (7.1)² - 2 × 4.1 × 7.1 × cos(45°)

∴ s² = 16.81 + 50.41 - 41.1677568

∴ s² = 26.0522432

→ Take √ for both sides

∴ s = 5.10413981

→ Round it to the nearest tenth

∴ s = 5.1 inches

∴ The length of s is 5.1 inches to the nearest tenth of an inch

3 0
3 years ago
Read 2 more answers
Find the solutions of the quadratic equation 2x2 + 6x - 2 = 0.
nekit [7.7K]

Sorry didn't know how to write the answer so I just took a screen shot of my answer for you!

6 0
2 years ago
A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t
Licemer1 [7]
If A(t) is the amount of salt in the tank at time t, then the rate at which this amount changes over time is given by the ODE

A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}

A'(t)+\dfrac1{100}A(t)=15

We're told that the tank initially starts with no salt in the water, so A(0)=0.

Multiply both sides by an integrating factor, e^{t/100}:

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}
\left(e^{t/100}A(t)\right)'=15e^{t/100}
e^{t/100}A(t)=1500e^{t/100}+C
A(t)=1500+Ce^{-t/100}

Since A(0)=0, we have

0=1500+C\implies C=-1500

so that the amount of salt in the tank over time is given by

A(t)=1500(1-e^{-t/100})

After 10 minutes, the amount of salt in the tank is

A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}
8 0
3 years ago
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