If R = 25 cm and S = 7 cm, what is the length of T?

Please help me question attached below:)))

atroni [7]

Answer:

D(5, 4), E(14, 7), M(9.5, 5.5)

Step-by-step explanation:

As AD = 1/4AB and DE ║ AC, the ratio CE/CB = 1/4, or CE = 1/4CB

Find the coordinates of D:

x = 1 + 1/4(17 - 1) = 1 + 4 = 5
y = 5 + 1/4(1 - 5) = 5 - 1 = 4

Find the coordinates of E:

x = 13 + 1/4(17 - 13) = 13 + 1 = 14
y = 9 + 1/4(1 - 9) = 9 - 2 = 7

Find the coordinates of the midpoint M of DE:

x = (5 + 14)/2 = 19/2 = 9.5
y = (4 + 7)/2 = 11/2 = 5.5

8 0

2 years ago

What is the correct letter? (Giving brainliest)

vampirchik [111]

Answer:

d is the answer

Step-by-step explanation:

perimeter of rect = 2(7+3) = 20

perimeter of square = 4 a =20

a = 5

3 0

2 years ago

A line intersects the points (-5, 1) and (-2,7). What is the slope of the line in simplest form? m=[?}

IrinaK [193]

Step-by-step explanation:

To find slope : y2- y1 / x2 - x1

we have ( -5 , 1 ) which is ( x1 , y1 )

and ( -2 , 7 ) is ( x2 , y2 )

so; 7 - 1 / -2 - (-5)

6/3

= 2

7 0

2 years ago

Which inequality is represented by the graph? y≥35x−1.5 y≤35x−1.5 y<35x−1.5 y>35x−1.5

Setler79 [48]

Answer:

y > 0.6x - 1.5

Step-by-step Explanation:

We need two points, to get to the equation of the graph.

Since we've got the following equation for two points (x1, y1), (x2, y2):-

$\boxed{ \mathsf{ \red{y - y_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} (x - x_{1}) }}}$

okay soo

I found two points that lie on this graph, not on the shaded region but yeah the dotted line which defines the graph.

one point is (0, -1.5) which lies on the y axis(the point where the dotted line touches the y axis)

other point is (2.5, 0) and this lies on the x axis

placing these points in the place of (x1, y1) and (x2, y2) in the above mentioned equation

$\mathsf{\implies y - ( - 1.5) = \frac{0 -( - 1.5)}{2.5 -0 } (x -0 )}$

you can take any one as (x1, y1) or (x2, y2).

so upon solving the above equation we get

$\mathsf{\implies (y + 1.5) = \frac{0 + 1.5}{2.5 } (x )}$

$\mathsf{\implies y + 1.5 = \frac{ 1.5}{2.5 } x }$

$\mathsf{\implies y + 1.5 = \frac{ \cancel{1.5}\:\:{}^3}{\cancel{2.5}\:\:{}^5 } x }$

$\mathsf{\implies y + 1.5 = \frac{ 3}{5 } x }$

multiplying both sides by 5

$\mathsf{5y + 7.5 = 3x}$

okay so this is the required equation of the dotted line

now we'll find the inequality

for this check whether the origin (0,0) lies under the shaded region or not

in this case it does

so

replacing x and y with 0

$\mathsf{\implies5(0) + 7.5 = 3(0)}$

$\mathsf{\implies0 + 7.5 = 0}$

this is absurd, 7.5 is not equal to 0 so we're gonna replace that equals sign with that of inequality

7.5 is greater than 0! so,

$\mathsf{\implies7.5 > 0}$

this goes for the whole equation, since we didnt swap any thing from left to right side of the equation or vice versa we can use this sign, to obtain the required inequality

$\mathsf{5y + 7.5 > 3x}$

dividing this inequality by 5, since there's no co-efficient in front of y in the given answers

we get

y + 1.5 > 0.6x

taking 1.5 to the RHS

that is the last option

5 0

3 years ago

What is 1/4 as a whole number?

wlad13 [49]

The whole number of $\dfrac{1}{4}$ is $\boxed{0.25}.$

Further explanation:

Always use the PEDMAS rule to solve the grouping of multiplication, addition, subtraction and division.

Here, P is parenthesis, E is exponents, M is multiplication, D is division, A is addition and S is subtraction.

Given:

The number is $\dfrac{1}{4}.$

Explanation:

The whole numbers is the series of numbers that starts from zero.

The natural numbers are those numbers that start from one.

The place values are only natural numbers.

The base ten systems represent the position of a place value.

The given number is $\dfrac{1}{4}.$

Consider the number $\dfrac{1}{4}$ as $\text{A}$ .

$A = \dfrac{1}{4}$

Divide 1 by 4 to obtain the whole of $\dfrac{1}{4}.$

$\begin{aligned}\text{A}&= \frac{1}{4}\\&= 0.25\\\end{aligned}$

The decimal expansion of $\dfrac{1}{4}$ is $0.25.$

The whole number of $\dfrac{1}{4}$ is $\boxed{0.25}.$

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Number system

Keywords: place value, base value, decimal expansion, whole number, natural number, division, multiplication, subtraction, solve the equation, solution, linear equation, line, decimal.

3 0

3 years ago

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