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=3x^2-6x+6x-12=3x(x-2)+6(x-2)=(3x+6)(x-2)

Other uses for X-rays and other types of radiation includecancertreatment. High-energyradiation in much higher doses than what is used for X-ray imaging may be utilized to help destroy cancerous cells and tumors by damaging their DNA.

Theraysgo through the skin and flesh easily, showing up as dark areas on the film, but with more difficulty through bone. They are slowed down and so these areas are much lighter.X-rayscan also beusedto kill cancer cells, but also kill healthy cells, so must beusedwith much care.

The most common form of X-ray used is X-ray radiography, which can be used to help detect or diagnose:

•Bone fractures •Infections (such aspneumonia)•Calcifications (likekidney stonesor vascular calcifications)•Some tumors•Arthritis in joints•Bone loss (such asosteoporosis)•Dental issues•Heart problems (such ascongestive heart failure)•Blood vessel blockages•Digestive problems•Foreign objects (such as items swallowed by children)

a=253,d=-5,n=20

a20=a+(n-1)d=253+(20-1)-5=253+(19)-5=253-95=168

168 is correct answer.

Triangle has three points

triangle

{2(-8-1)-1(4-3)+3(6+1)}={2(-9)-1+3(7)}={-18-1+21}={1}; x=8; y=4, z=0

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not answer

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Option b is correct12y=15xx/y=15/12Option d is correct12/15x=yx/y=15/12Option a3x+3y=27x-27y-24x=-30yx/y=15/12Hence option d is incorrect

can you please do it in paper

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May this help u and try itself to do it again

May this help u and try itself to do again

5 is the correct answer

Given: ABAC and DEFE such that,

AB = DE and BF = CD

To prove: AC = EF

Proof:

InABC, we have,

BC = BF + FC

and, inDEF

FD = FC + CD

But, BF = CD [Given]

So, BC = BF + FC

and, FD = FC + BF

BC = FD

So, inABC andDEF, we have,

BAC =DEF = 90o[Given]

BC = FD [Proved above]

AB = DE [Given]

Thus, by Right angle-Hypotenuse-Side criterion of congruence, we have

ABCDEF

The corresponding parts of the congruent triangle are equal.

So, AC = EF [c.p.c.t]

AE =BCD [Proved above]

Thus by Angle-Side-Angle criterion of congruence, we have

BCDBBAE

The corresponding parts of the congruent triangles are equal.

So, CD = AE [Proved]

In the figure, OPT is a right angled triangle, right angled a T (As PT is a tangent).So, let OT and OA be r.Thus,r² + PT² = OP²or, r² + 6² = (r+3)²or, r² + 36 = r² + 6r +9or, 6r + 9 = 36 (Cutting off r² from both sides)or, 6r =27or, r = 27/6 = 9/2 = 4.5cm

you are write

Area =length*breadth2450=2b*b1225=b^2b=35mPerimeter=2*(l+b)=2*(70+35)=2*105=210moption d is correct

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none of these is the answers 4.681 is come

16.123+4.516+x=25.32=20.639+x=25.32=x=25.32-20.639x=4.618

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Try to solve the answer for trigonomentry formula

DE || BC Area (ADE) : Area (ABC) = (AD:DB)(AD:DB) = 9:25

[29-(-2){6-(7-3)}] ÷[3×{5+(-3)×(-2)}]

= [29+2(6-4)]÷[3×{5+6}]= [29+4]÷[3×11]= 33/33= 1

5:3=5/3=1.6729:7=29/7=4.1419:2=19/2=9.532:9=32/9=3.55so 19:2 is maximum

Additional area grazed=(π.23² - π.16²)/4=214.305 m²

As DE || BC

Use Basic Proportionality Theorem

DE/BD = AE /EC

2.4/BD = 3.2/4.8

BD = (2.4 × 4.8)/3.2

BD = 3.6 cm

c) is correct

400×75/100

400 ×75/100= 300

75 % of Rs 400,75 × 400 ÷ 100 .(Zero zero cut)75 × 4 Rs 300.Therefore, The 75 % of Rs 400 is Rs 300.

i)by differentiating we will get slope of tangentslope of normal=1/slope of tangent=1/0at x=(pi)/6 , y=2y=mx+cy=(1/0)x+cx=0 is the normal to the given curveii)same as 1

ΔABC and ΔADC are two equilateral triangles on a common base AC.

In triangle ABC, AB = AC = BC and in triangle ADC, AD = AC = DC

From this we get two triangles ABC and ADC are Equilateral Triangles with same side.

The equilateral triangle have same angle which is equal to 60°

Here AC is the common base, We get a quadrilateral ABCD.

<A= 60°+60 °=120°, <B=60°, <C=60°+60° =120° and <D=60°

Also AB=BC=CD=AD

All the sides of this quadrilateral are equal , so it is a rhombus.