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IrinaVladis [17]
3 years ago
11

WILL MARK BRAINLIEST HELP ASAP

Mathematics
2 answers:
eduard3 years ago
8 0
Just substitute the x and you get 8 times 4 minus 3
OLga [1]3 years ago
4 0
I’m not sure if this is right. But I think it’s 29

x=4
8(4) -3
32-3=29
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Use four unit multipliers to convert 36 square yards to square inches. (Use two unit multipliers to go from square yards to squa
Tom [10]
<span>1 sq yd = 36^2 sq in = 1296 sq in
so 170 sq in = 170/1296 sq yd


1 yd = 3 ft
1 ft = 12 in
170 sq in [ft/12 in] [ft/12 in] [yd/3 ft][yd/3 ft</span>
7 0
3 years ago
A right triangle has an area of 21 m2. The dimensions of the triangle are increased by a scale factor of 4. What is the area of
Zinaida [17]

Answer:

168 meters squared

Step-by-step explanation:

The area of the triangle is A= 1/2 b*h. Here this area is 21 meters squared. The sides of the triangle have been multiplied by 4. This means, if the base was b, it is now 4b. This also means, if the height was h, it is now 4h. So the area is A= 1/2 4b*4h = 1/2 (16bh) = 8bh.

So the area of the larger triangle will be 8 times bigger.

21*8 = 168

3 0
4 years ago
Which are perfect trinomials?
Natalija [7]

Answer:

A perfect square trinomial is the square of a binomial. It follows a pattern when it is factored, so that the first and last terms are perfect squares of monomials and the middle term is twice their product

Step-by-step explanation:

5 0
3 years ago
What is x if 389/x=47
Agata [3.3K]
<span>389/x=47

To solve this, you must isolate x on one side.
First multiply both sides by x.

x(389/x) = x(47)
389 = 47x

Now divide both sides by 47 to find x.
389/47 = x

X is 389/47.</span>
7 0
3 years ago
The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion s
kotykmax [81]

Answer:

(i) [1,2]

V_a_v_g=\frac{s(2)-s(1)}{2-1} =\frac{2-(-2)}{1} =4cm/s

(ii) [1,1,1]

V_a_v_g=\frac{s(1.1)-s(1)}{1.1-1} =\frac{-3.447-(-2)}{0.1} =-14.47cm/s

(iii)

V_a_v_g=\frac{s(1.01)-s(1)}{1.01-1} =\frac{-2.156-(-2)}{0.01} =-15.6cm/s

(iv)

V_a_v_g=\frac{s(1.001)-s(1)}{1.001-1} =\frac{-2.016-(-2)}{0.001} =-15.698cm/s

(b)

\frac{ds}{dt} \left \{  {t=1}} \right =-15.7076327\approx-15.7cm/s

Step-by-step explanation:

s(t)=5sin(\pi t)+2cos(\pi t)

First. Let's find the displacement for every given time:

t=1s

s(1)=5sin(\pi)+2cos(\pi)=-2cm=-0.02m

t=2s

s(2)=5sin(\pi *2)+2cos(\pi *2)=2cm=0.02m

t=1.1s

s(1.1)=5sin(\pi *1.1)+2cos(\pi *1.1)\approx -3.447cm\approx-0.03447m

t=1.01s

s(1.101)=5sin(\pi *1.101)+2cos(\pi *1.101)\approx -2.156cm\approx-0.02156m

t=1.001s

s(1.001)=5sin(\pi *1.001)+2cos(\pi *1.001)\approx -2.016cm\approx-0.02016m

Now, the average velocity can be found as:

V_a_v_g=\frac{S_f-S_i}{t_f-t_i}

Where:

S_f=Final\hspace{3}displacement\\S_i=Initial\hspace{3}displacement\\t_f=Final\hspace{3}time\\t_i=Initial\hspace{3}time\\

Hence:

(i) [1,2]

V_a_v_g=\frac{s(2)-s(1)}{2-1} =\frac{2-(-2)}{1} =4cm/s

(ii) [1,1,1]

V_a_v_g=\frac{s(1.1)-s(1)}{1.1-1} =\frac{-3.447-(-2)}{0.1} =-14.47cm/s

(iii)

V_a_v_g=\frac{s(1.01)-s(1)}{1.01-1} =\frac{-2.156-(-2)}{0.01} =-15.6cm/s

(iv)

V_a_v_g=\frac{s(1.001)-s(1)}{1.001-1} =\frac{-2.016-(-2)}{0.001} =-15.698cm/s

(b) In order to estimate the instantaneous velocity of the particle for t=1, we need to find its derivative:

\frac{ds}{dt} \left \{  {t=1}} \right.  =5\pi cos(\pi t) -2\pi sin(\pi t)=5\pi cos(\pi (1)) -2\pi sin(\pi (1)) \\\\\frac{ds}{dt} \left \{  {t=1}} \right =-15.7076327\approx-15.7cm/s

4 0
3 years ago
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