All categories

2 years ago

Solv the following inequality. p – 4 < -11

Mathematics

1 answer:

bogdanovich [222]2 years ago

6 0

Answer:

p < -7

Step-by-step explanation:

.................

You might be interested in

Please help will give brainliest

elena-14-01-66 [18.8K]

Answer:

157 cm^2

Step-by-step explanation:

1) find the radius

radius = side of the square /2 = 10 /2 = 5 cm

2) notice that four semicircles are equals to two circles

3) find the area of the shaded region

A = radius^2 x pi x 2 = 5^2 x pi x 2 = 157 cm^2 (round the nearest whole number)

8 0

2 years ago

Please write an expression if u do all correct I give brainliest

Snezhnost [94]

Answer:

1.2 + 2.3

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The figure below has a point marked with a large dot.

defon

Answer:

( 5 , - 3 ) -------> ( 3 , 5 )

Step-by-step explanation:

4 0

2 years ago

Find x and the measure of each side. Round to two decimal places, if necessary.

natta225 [31]

The values of the triangles are as follows:

x = 7 units
GH = 21 units
HI = 21 units
GI = 12 units

<h3>How to find angles and side of a triangle?</h3>

The triangle is an isosceles triangle because two sides and angles are equal. Therefore,

4x - 7 = 2x + 7

4x - 2x = 7 + 7

2x = 14

x = 14 / 2

x = 7 units

GH = 4(7) - 7 = 28 - 7 = 21

HI = 2(7) + 7 = 14 + 7 = 21

GI = 7 + 5 = 12

learn more on triangle here: brainly.com/question/21279088

#SPJ1

3 0

1 year ago

Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and pro

Neporo4naja [7]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

★ Sanya has a piece of land which is in the shape of a rhombus.

★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

★ Perimeter of land = 400 m.

★ One of the diagonal = 160 m.

${\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}$

★ Area each of them [son and daughter] will get.

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

Let, ABCD be the rhombus shaped field and each side of the field be $x$

[ All sides of the rhombus are equal, therefore we will let the each side of the field be $x$ ]

Now,

• Perimeter = 400m

$\longrightarrow \tt AB+BC+CD+AD=400m$

$\longrightarrow \tt x + x + x + x=400$

$\longrightarrow \tt 4x=400$

$\longrightarrow \tt \: x = \dfrac{400}{4}$

$\longrightarrow \tt x= \red{100m}$

$\therefore$ Each side of the field = 100m.

Now, we have to find the area each [son and daughter] will get.

So, For $\triangle$ ABD,

Here,

• a = 100 [AB]

• b = 100 [AD]

• c = 160 [BD]

$\therefore \tt Simi \: perimeter \: [S] = \boxed{ \sf \dfrac{a + b + c}{2} }$

$\longrightarrow \tt S = \dfrac{100 + 100 + 160}{2}$

$\longrightarrow \tt S = \cancel{ \dfrac{360}{2}}$

$\longrightarrow \tt S = 180m$

Using herons formula,

$\star \tt Area \: of \: \triangle = \boxed{\bf{{ \sqrt{s(s - a)(s - b)(s - c) } }}} \star$

where

• s is the simi perimeter = 180m

• a, b and c are sides of the triangle which are 100m, 100m and 160m respectively.

Putting the values,

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{180(180 - 100)(180 - 100)(180 - 160) }$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{180(80)(80)(20) }$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{180 \times 80 \times 80 \times 20 }$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{9 \times 20 \times 20 \times 80 \times 80}$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \tt \sqrt{ {3}^{2} \times {20}^{2} \times {80}^{2} }$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = 3 \times 20 \times 80$

$\longrightarrow \tt Area_{ ( \triangle \: ABD)} = \red{ 4800 \: {m}^{2} }$

Thus, area of $\triangle$ ABD = 4800 m²

As both the triangles have same sides

So,

Area of $\triangle$ BCD = 4800 m²

Therefore, area each of them [son and daughter] will get = 4800 m²

${\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}$

★ Figure in attachment.

${\underline{\rule{290pt}{2pt}}}$

7 0

1 year ago

Read 2 more answers