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3 years ago

Please please help me

Mathematics

1 answer:

CaHeK987 [17]3 years ago

3 0

Answer:

2 or 1

Step-by-step explanation:

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Find the sum: –1.54 + 5.093 ...?

Dvinal [7]

3.553 use a calculator

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3 years ago

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Find the midpoint of the segment with the given endpoints. (-7, 10) and (4,-9)

posledela

Answer:

(-20 - 10) / (-12 - (-10))

-30 / -2

30/2

now the midpoint would be half the two points

30/2 ÷ 2/2 = 15/1

So add 15 to -20 (or subtract 15 from 10) for the y, and add 1 to -12 (or subtract 1 from -10) for the x.

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3 years ago

Find the equation of the line that passes through (1,4) and is parallel to 3x+y−1=0. Leave your answer in the form y=mx+c

Leto [7]

Answer:

y = - 3x + 7

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

3x + y - 1 = 0 ( subtract 3x - 1 from both sides )

y = - 3x + 1 ← in slope- intercept form

with slope m = - 3

Parallel lines have equal slopes, thus

y = - 3x + c ← is the partial equation

To find c substitute (1, 4) into the partial equation

4 = - 3 + c ⇒ c = 4 + 3 = 7

y = - 3x + 7 ← equation of line in form y = mx + c

7 0

3 years ago

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LOGARITHMS, CONIC SECTIONS, SEQUENCES

pochemuha

Answer:

(a) 40 milligrams

(b) 1.21 milligrams

Step-by-step explanation:

Exponential Function

It's commonly used to model real situations in life, like this one proposed in the question where the number of milligrams of a drug D(h) in a patient's bloodstream h hours after the drug is injected is modeled by the following function.

$D(h) =40e^{-0.5h}$

(a) The initial amount injected to the patient can be found when h=0

$D(0) =40e^{-0.5(0)}=40e^{0}=40$

Initial amount: 40 milligrams

(b) Let's plug h=7

D(7) =40e^{-0.5(7)}=40e^{-3.5}=1.21

The amount in the bloodstream after 7 hours is 1.21 milligrams

4 0

4 years ago

A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$

$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$

$$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}$

$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$

$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$

$$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square$

8 0

2 years ago