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balu736 [363]
3 years ago
11

Three triangles are shown on the centimetre grid. A, B, C. A\ Which triangle has the largest area? B\ Work out the area of this

triangle. Give your answer a.S a decimal.

Mathematics
2 answers:
ahrayia [7]3 years ago
8 0

Answer:

A) The three triangles have the same area

B) 10 square units

Step-by-step explanation:

In the figure attached, the triangles are shown.

A) The three triangles have the same area because all have the same base and height (see figure)

B) Area of a triangle = (1/2)*base*height

Area of a triangle = (1/2)*4*5 = 10 square units

kolbaska11 [484]3 years ago
4 0

Answer:

A) = C

B) = 4.5 cm^2

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Three quarter of the students running a 100-yard race finished with an average time of 16 seconds. The remaining 25% of students
iren [92.7K]

Answer:

15 seconds

Step-by-step explanation:

Because the split id 25% and 75%, we could create another average pretending that there are four kids, one who ran in 12 seconds, and three who ran in 16.

Equation for averages: (a₁ + a₂ + a₃ + ... a_{n})/ n

Plug in:<em> (12 + 16 + 16 + 16)/4</em>

Add: 60/4

Divide: 15 seconds

5 0
3 years ago
Y<br> 45°<br> X 100°<br> y = [ ? 1°
deff fn [24]

Step-by-step explanation:

x+100°=180°{being straight angle}

x=180-100

x=80°

again,

45°+x+y=180°{sum of angle of triangle}

45+80+y=180

y=180-125°

y=55°

hope it helps.

8 0
3 years ago
Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview
mario62 [17]

Answer:  The required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

Step-by-step explanation:   We are given to solve the following differential equation :

5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)

Let us consider that

y=e^{mt} be an auxiliary solution of equation (i).

Then, we have

y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.

Substituting these values in equation (i), we get

5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.

So, the general solution of the given equation is

y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.

Differentiating with respect to t, we get

y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.

According to the given conditions, we have

y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)

and

y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.

From equation (ii), we get

B=\dfrac{7}{3}.

Thus, the required solution is

y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.

7 0
3 years ago
Find the common ratio 27,9,3,1,1/3
LenKa [72]

Answer:

1/3

Step-by-step explanation:

To find the common ratio, take the second term and divide by the first term

9/27 = 1/3

Check by dividing the third term by the second

3/9 = 1/3

and so forth

1/3 = 1/3

etc

The common ratio is 1/3

8 0
3 years ago
Read 2 more answers
In the context of relationship between variables, increases in the values of one variable are accompanied by increases in the va
garik1379 [7]

Answer:

A. Positive linear.

Step-by-step explanation:

We have that both variables increases, then we have a positive relation. A curvilinear option can not be possible because with this option in some regions could happen that when one variable increases the other one decreases. The negative linear relation can no be because with this option when one variable increases the other decreases. A non linear option is the same as a curvilinear option then can not be possible. Then the best option is a positive linear relationship.

7 0
2 years ago
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