All categories

1 year ago

Mary is four years older than Sue. If the sum of their ages is 16, how old are Mary and Sue?

Mathematics

2 answers:

dusya [7]1 year ago

6 0

Answer:

Sue is 6 and Mary is 10

Step-by-step explanation:

let Sue's age be x then Mary is x + 4 and the sum of their ages is

x + x + 4 = 16

2x + 4 = 16 ( subtract 4 from both sides )

2x = 12 ( divide both sides by 2 )

x = 6

x + 4 = 6 + 4 = 10

then Mary is 10 and Sue is 6

DedPeter [7]1 year ago

4 0

Answer: Mary 10 Sue 6

Step-by-step explanation:

You might be interested in

What’s the answer plzzz hurry

ExtremeBDS [4]

I can’t see it it too blurry

7 0

3 years ago

Read 2 more answers

14 POINTS 7x - 3 = 18

Svetlanka [38]

Subtract 3 first so it would be 7x=15
then divide by 7 so x would be approximately 2.14

5 0

3 years ago

Read 2 more answers

Some teenagers collected trash for a beach cleanup. The data for the number of pounds of trash collected by each teenager are sh

Mumz [18]

Answer:

4.66

Step-by-step explanation:

We are given the following data

26, 26, 21, 22, 20, 25, 35

Step 1

We find the Mean

Mean = 26+ 26+ 21+ 22+20+25+ 35/7

= 175/7

= 25

Step 2

We find the Standard Deviation

The formula for Standard deviation that we would use is the Population Standard deviation, this is because the data given above is the data of the number of pounds of trash collected by each teenager

= √(x - Mean)²/n

= √(26 - 25)²+( 26 - 25)²+ (21 - 25)²+ (22 - 25)² +(20 - 25)²+(25 - 25)²+ (35 - 25)²/7

= √152/7

= √21.7142857

= 4.65985898

Approximately = 4.66

5 0

3 years ago

Read 2 more answers

Surface integrals using an explicit description. Evaluate the surface integral \iint_{S}^{}f(x,y,z)dS using an explicit represen

Jobisdone [24]

Parameterize $S$ by the vector function

$\vec r(x,y)=x\,\vec\imath+y\,\vec\jmath+f(x,y)\,\vec k$

so that the normal vector to $S$ is given by

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=\left(\vec\imath+\dfrac{\partial f}{\partial x}\,\vec k\right)\times\left(\vec\jmath+\dfrac{\partial f}{\partial y}\,\vec k\right)=-\dfrac{\partial f}{\partial x}\vec\imath-\dfrac{\partial f}{\partial y}\vec\jmath+\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}\right\|=\sqrt{\left(\dfrac{\partial f}{\partial x}\right)^2+\left(\dfrac{\partial f}{\partial y}\right)^2+1}$

In this case, the normal vector is

$\dfrac{\partial\vec r}{\partial x}\times\dfrac{\partial\vec r}{\partial y}=-\dfrac{\partial(8-x-2y)}{\partial x}\,\vec\imath-\dfrac{\partial(8-x-2y)}{\partial y}\,\vec\jmath+\vec k=\vec\imath+2\,\vec\jmath+\vec k$

with magnitude $\sqrt{1^2+2^2+1^2}=\sqrt6$ . The integral of $f(x,y,z)=e^z$ over $S$ is then

$\displaystyle\iint_Se^z\,\mathrm d\Sigma=\sqrt6\iint_Te^{8-x-2y}\,\mathrm dy\,\mathrm dx$

where $T$ is the region in the $x,y$ plane over which $S$ is defined. In this case, it's the triangle in the plane $z=0$ which we can capture with $0\le x\le8$ and $0\le y\le\frac{8-x}2$ , so that we have

$\displaystyle\sqrt6\iint_Te^{8-x-2y}\,\mathrm dx\,\mathrm dy=\sqrt6\int_0^8\int_0^{(8-x)/2}e^{8-x-2y}\,\mathrm dy\,\mathrm dx=\boxed{\sqrt{\frac32}(e^8-9)}$

5 0

3 years ago

Which method can you use to prove that the triangles are congruent

antoniya [11.8K]

Answer:

ΔASA ≅ ΔASA

Step-by-step explanation:

4 0

3 years ago