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3 years ago

5

Help needs to be done in 10 mins

1 answer:

Igoryamba3 years ago

4 0

Answer:

Step-by-step explanation:

For The First Box Across From -4 Would Be 19

For Second Box Across From 3 Would Be -2

For Third Box Across From 2 Would Be 1

For The Fourth Box Across From 16 Would Be 25

All You Have To Do Is Multiply The Number By The X In Each Box

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Someone please help me??

lapo4ka [179]

The standard equation of a circle with centre (xc,yc) and radius R is given by

$(x-x_c)^2+(y-y_c)^2=R^2$

Substituting
centre (xc,yc) = (4,-3)
R=2.5

The equation is therefore

$(x-x_c)^2+(y-y_c)^2=R^2$
$(x-4)^2+(y-(-3))^2=2.5^2$
$(x-4)^2+(y+3)^2=2.5^2$ or
(x-4)^2+(y+3)^2=6.25

7 0

3 years ago

If the value of a is 8 in f(x)=ax^2, what does the graph look like?

viktelen [127]

Step-by-step explanation:

the answer is in the image above

6 0

3 years ago

Show that y=cos(t)y=cos(t) is a solution to (dydt)2=1−y2(dydt)2=1−y2. Enter your answers below in terms of the independent varia

GrogVix [38]

Answer:

$(\frac{dy}{dt})^2=sin^2t$

Step-by-step explanation:

$y=cost$

DE : $(\frac{dy}{dt})^2=1-y^2$

If y is a solution of given DE then it satisfied the DE.

Differentiate w.r.t t

$\frac{dy}{dt}=-sint$

Using the formula

$\frac{d(cosx)}{dx}=-sinx$

LHS: $(\frac{dy}{dt})^2=(-sint)^2=sin^2t$

RHS

$1-y^2=1-cos^2t=sin^2t$

By using the formula

$sin^2t=1-cos^2t$

LHS=RHs

Hence, y is a solution of given DE

$(\frac{dy}{dt})^2=sin^2t$

7 0

3 years ago

Evaluate sum r=1 ^ infty 1 r(r+2)^ prime

sasho [114]

1^infinity = 1

Thus :

r = 1

______________________

r ( r + 2 )^prime =

1 × ( 1 + 2 )^1 =

1 × 3^1 =

1 × 3 =

3

3 0

3 years ago

Read 2 more answers

Please can someone help me with calculating area and finding volume to cubes/rectangular prisms

nadezda [96]

When u find the volume , u multiply the length and width and height together

6 0

3 years ago