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3 years ago

Please help ASAP ASAP

Mathematics

2 answers:

viva [34]3 years ago

5 0

Answer:X=110

Step-by-step explanation:

When you divide 196 by 20 you get 9.8 so when you divide 1078 by 9.8 you get 110

GuDViN [60]3 years ago

3 0

I think that the answer is 110=x

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I do not know how to this question and i need help please

maxonik [38]

Answer:

---

Step-by-step explanation:

I know this probably won't help but, your finding rate, To solve for speed or rate use the formula for speed, s = d/t which means speed equals distance divided by time.

6 0

3 years ago

The point (1, –5) is reflected across the y-axis.

MArishka [77]

Answer:

(1, 5)

Step-by-step explanation:

Reflecting across the y-axis is the same thing as adding a negative to the y-value.

double negative 5 is the same thing as positive 5 so

(1, 5) is your answer

7 0

3 years ago

1) 14 units2 2) 16 units2 3) 17 units2 4) 19 units2

horrorfan [7]

Answer:

<h2>4th option: 19 units2 </h2>

Step-by-step explanation:

<h3>The given figure can be divided as (i) upper rectangle, (ii) lower rectangle </h3>

Then,

(i) Breadth of upper rectangle = 2 units

Length of upper rectangle = 8 - 1 = 7 units

Therefore, 

Area of the upper rectangle = 7 × 2 = 14 units2

(ii) Breadth of lower rectangle = 1 unit

Length of lower rectangle = 5 units

Therefore, 

Area of the lower rectangle = 5 × 1 = 5 units2

Since,

Area of the figure is = Area of upper rectangle + Area of lower rectangle 

= 14 units2 + 5 units2

= 19 units2 (Ans) (4th option)

7 0

2 years ago

The working out for 637 divided by 13

Sedbober [7]

I thing the answer you are looking for is 49

5 0

3 years ago

Read 2 more answers

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

ivolga24 [154]

Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

6 0

3 years ago